# A Path to Fractional Integral Representations of Some Special Functions

Common Topics: function, integrals, dirichlet, integral, fractional

## Introduction

This bit is what new thing you can learn reading this:) As for original content, I only have hope that the method of using the sets

$$C_N^n: = \left\{ { \vec x \in {\mathbb{R}^n}|{x_i} \ge 0\forall i,\sum\limits_{k = 1}^n {x_k^{2N}} < n – 1 } \right\}$$

and Dirichlet integrals to evaluate certain integrals of the type

$$\mathop {\lim }\limits_{N \to \infty } \int\limits_{C_N^n} {f\left( {\vec x} \right)d\mu } = \int\limits_{{{\left[ {0,1} \right]}^n}} {f\left( {\vec x} \right)d\mu }$$

might be original material as I have never seen it my reading.

### Summary

The main purpose of this paper is to derive the formulas in Sections 4 and 5. Section 4 hold n-fold iterated integral representations of some special functions (where n is a positive integer), though somewhat dense, all the material up to this and including Section 3 is just advanced Calc 3 level material; Sections 4&5 are the analysis content, section 5 contains fractional integrals as analytic continuations of the previous section’s formulas on the variable n which gets continued to a complex-valued parameter.

Just what is a fractional integral? Ever hear of n-fold integrals? What would it mean to allow complex numbers for the order of integration? This is what fractional integrals are, a generalization of integration.

#### Section 1

Gives a glimpse of the gamma and beta functions. As for special functions, the gamma function is the least special function and should be the first special function one meets: it is the analytic continuation of the factorial. It arises in many venues: mathematical and engineering to be sure, and others as well. The beta function is defined in terms of gamma functions.

#### Section 2

It Covers Dirichlet Integrals which are handy for evaluating certain n-dimensional integrals over a general class of domains in terms of the gamma function.

#### Section 3

The purpose of this section is not readily apparent, trust me we’ll need this in section 4 and it works like magic in combination with the Dirichlet integrals we studied in the last section, this section is dedicated mostly to defining a sequence of sets that point-wise converges to an orthotope (the sets ##C_N^n## above) which will be used in section 4 to evaluate certain multiple integrals over the unit hypercube.

#### Section 4

Involves some formulas for certain special functions represented as n-fold iterated integrals over the unit hypercube evaluated in a unique fashion. These special functions are the Lerch Transcendent, Legendre Chi, Polygamma, Polylogarithm of Order n, Hurwitz Zeta, Dirichlet Beta, Dirichlet Eta, and the Dirichlet Lambda functions (all of these depend on a positive integer n being the order of integration).

#### Section 5

Expands the previous section’s special functions represented as multiple integrals to fractional integrals which provide analytic continuations of the prior section’s identities to complex orders of integration (n is continued to a complex-valued variable,) in particular, the Hadamard fractional integral operator is employed to this end.

## The Integrals of Dirichlet

The proofs for Dirichlet Integrals have been allowed to follow as corollaries of the more general theorem of Louisville in Appendix A, and this has been done to drastically reduce the reading chore. In this section instead of proofs we simply state the the two most used corollaries in the remainder of this work.

Dirichlet integrals as I learned them from an Advanced Calculus book are just that formula evaluating the integral to Gamma functions, they are not a type of integral like Riemann integral, more just a formula that would go on a table of integrals. Content is the 4+-dimensional version of volume (some writers use hypervolume instead of content).

A result due to Dirichlet is given by

#### Corollary 2.2: Dirichlet Integrals (Modified Domain 1)

If ##t,{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_t^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq t} } \right\}##, then

$$\iint {\mathop \cdots \limits_{V_t^n} \int {\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {t^{\sum\limits_{p = 1}^n {\frac{{{\gamma _p}}}{{{\beta _p}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}$$

#### Corollary 2.4: Content Integral

If ##{\alpha _p},{\beta _q} > 0\forall p,q## and ##{V^n}: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq 1} } \right\}##, then

##{\text{Content}}\left( {{V^n}} \right): = \iint {\mathop  \cdots \limits_{{V^n}} \int {d{z_n} \ldots d{z_2}d{z_1}} } = {{\prod\limits_{q = 1}^n {\left[ {\frac{{{\alpha _p}}}{{{\beta _q}}}\Gamma \left( {\frac{1}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{{\alpha _p}}}{{{\beta _q}}}\Gamma \left( {\frac{1}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{{\beta _k}}}} } \right)}}##. Where content refers to hypervolume when ##n \geq 4##.

## A bit of Geometry

### The Orthotope

The orthotope is the generalization of the rectangular parallelepiped to ##{\mathbb{R}^n}##. Let ##{b_k} > 0\forall k \leq n## and consider the orthotope with polytope vertices of the form ##\left( { \pm {b_1}, \pm {b_2}, \ldots , \pm {b_n}} \right)## orientated with facet-centered axes determined by the set ##{P^n}: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\left| {{x_k}} \right| \leq {b_k}\forall k \leq n} \right\}##. It will be convenient to develop a definition of the orthotope which may seem overly complicated but will make good use of it in the next section. Let ##S_N^n: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N}}}  \leq n} \right\}##. Then we proceed to prove by bivariate induction on n and N that

#### Lemma 3.1: Nesting Property of ##S_N^n##

If ##S_N^n: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N}}}  \leq n} \right\}##, then ##S_{N + 1}^n \subset S_N^n##.

##### Proof:

By bivariate induction on n and N (this has three parts)[1]:

First, we define the statement to be proved ##P\left( {n,N} \right):S_{N + 1}^n \subset S_N^n##.

(i) Prove the base case ##P\left( {a,b} \right)## where ##a,b \in {\mathbb{Z}^ + }## are the base or smallest values of a and b such that ##P\left( {a,b} \right)## holds. For this problem that means proving ##P\left( {2,1} \right):S_2^2 \subset S_1^2##. We have ##S_2^2: = \left\{ {\left( {{x_1},{x_2}} \right) \in {\mathbb{R}^2}|\sum\limits_{k = 1}^2 {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}}  \leq 2} \right\}## implies the following inequality holds ##\sum\limits_{k = 1}^2 {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}}  \leq 2##. Now to work given inequality down we will use calculus to find the values that maximize the constraint ##f\left( {{x_1},{x_2}} \right) = \sum\limits_{k = 1}^2 {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}}  \leq 2## and the ad hoc constraint ##g\left( {{x_1},{x_2}} \right) = \tfrac{{{x_1}}}{{{b_1}}} + \tfrac{{{x_2}}}{{{b_2}}} = c## for some positive constant c (which is certainly true). We proceed by the method of Lagrange multipliers

$$\nabla f = \left\langle {4\tfrac{{x_1^3}}{{b_1^4}},4\tfrac{{x_2^3}}{{b_2^4}}} \right\rangle = \lambda \left\langle {\tfrac{1}{{{b_1}}},\tfrac{1}{{{b_2}}}} \right\rangle = \lambda \nabla g \Rightarrow \lambda = 4{\left( {\tfrac{{{x_1}}}{{{b_1}}}} \right)^3} = 4{\left( {\tfrac{{{x_2}}}{{{b_2}}}} \right)^3} \Leftrightarrow {x_2} = \tfrac{{{b_2}}}{{{b_1}}}{x_1}$$

Plug this relation into the former inequality, namely

##f\left( {{x_1},\tfrac{{{b_2}}}{{{b_1}}}{x_1}} \right) = {\left( {\tfrac{{{x_1}}}{{{b_1}}}} \right)^4} + {\left( {\tfrac{{{x_1}}}{{{b_1}}}} \right)^4} \leq 2 \Rightarrow {\left( {\tfrac{{{x_1}}}{{{b_1}}}} \right)^2} \leq 1 = :{\max _{1 \leq i \leq 2}}\left\{ {{{\left( {\frac{{{x_i}}}{{{b_i}}}} \right)}^2}} \right\}##

where ##{\max _{1 \leq i \leq n}}\left\{ {{a_i}} \right\}: = \max \left\{ {{a_1},{a_2}, \ldots ,{a_n}} \right\}##. Checking to see if ##S_2^2 \subset S_1^2## by plugging in maximal values we get ##\sum\limits_{k = 1}^2 {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^2}}  \leq 2{\max _{1 \leq i \leq 2}}\left\{ {{{\left( {\frac{{{x_i}}}{{{b_i}}}} \right)}^2}} \right\} = 2##, and hence ##S_2^2 \subset S_1^2## the base case is proven to hold.

(ii) Now we prove induction over n: assume that ##P\left( {n,b} \right)## holds for some fixed ##n \in {\mathbb{Z}^ + }## and where b is as in part (i). Then we need to prove that ##P\left( {n + 1,b} \right)## is therefore true. For this problem that means proving ##P\left( {n,1} \right):S_2^n \subset S_2^n \Rightarrow P\left( {n + 1,1} \right):S_2^{n + 1} \subset S_2^{n + 1}##.

Assume $$P\left( {n,1} \right) :S_2^n: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}} \leq n} \right\} \subset \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}| \sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^2}} \leq n} \right\} = :S_1^n$$ (it turns out we will not need this assumption to prove what needs to be proven, just some simple calculus). Now let’s look at

$$P\left( {n + 1,1} \right):S_2^{n + 1}: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}|\sum\limits_{k = 1}^{n + 1} {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}} \leq n + 1} \right\} \subset \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}|\sum\limits_{k = 1}^{n + 1} {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^2}} \leq n + 1} \right\} = :S_1^{n + 1}$$

Let us just use calculus to maximize the function ##F\left( {\vec x} \right): = \sum\limits_{k = 1}^{n + 1} {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^2}} ## subject to the constraint  ##G\left( {\vec x} \right): = \sum\limits_{k = 1}^{n + 1} {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^4}}  \leq n + 1 ##. Using the method of Lagrange Multipliers again we have (in component form): $${\left( {\nabla F} \right)_i} = 2\tfrac{{{x_i}}}{{b_i^2}} = 4\lambda \tfrac{{x_i^3}}{{b_i^4}} = \lambda {\left( {\nabla G} \right)_i} \Rightarrow \left( {{x_i} = 0} \right) \vee \left[ {\left( {{x_i} \ne 0} \right) \wedge \left( {\lambda = \tfrac{1}{2}{{\left( {\tfrac{{{b_i}}}{{{x_i}}}} \right)}^2}} \right)} \right]$$ $$\Rightarrow {z^2}: = {\left( {\tfrac{{{x_i}}}{{{b_i}}}} \right)^2} = {\left( {\tfrac{{{x_j}}}{{{b_j}}}} \right)^2}\,\forall i \ne j \leq n + 1$$ is when the maximum occurs because obviously the zero was a minimum. Plugging into the constraint to get a bound for ##{z^2}## gives ##\sum\limits_{k = 1}^{n + 1} {{z^4}}  = \left( {n + 1} \right){z^4} \leq n + 1 \Rightarrow {z^2} \leq 1## and then to find the actual maximum we get ##F\left( {\vec x} \right) \leq \sum\limits_{k = 1}^{n + 1} {{z^2}}  = \left( {n + 1} \right){z^2} = n + 1## and hence ##S_2^{n + 1} \subset S_1^{n + 1}##.

(iii) Now for the induction on N step: Assume ##P\left( {n,N} \right)##  holds for some fixed N. We must prove that ##P\left( {n,N + 1} \right)## is therefore true. Notably we will not use the assumption in this step. We have ##S_{N + 2}^n: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N + 4}}}  \leq n} \right\}##. From this the inequality ##\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N + 4}}}  \leq n## holds for fixed N, so we should maximize ##\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N + 2}}}##  subject to that constraint: proceeding by Lagrange Multipliers method in component form we have ##\left( {2N + 2} \right)\tfrac{{x_k^{2N + 1}}}{{b_k^{2N + 2}}} = \left( {2N + 4} \right)\lambda \tfrac{{x_k^{2N + 3}}}{{b_k^{2N}}} \Rightarrow \frac{{N + 1}}{{N + 2}}\tfrac{{b_k^2}}{{x_k^2}} = \lambda  \Rightarrow w: = \tfrac{{x_k^2}}{{b_k^2}} = \tfrac{{x_j^2}}{{b_j^2}}\forall j \ne k \leq n## now plug this value into the constraint to get ##\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N + 4}}}  = \sum\limits_{k = 1}^n {{w^{2N + 4}}}  = n{w^{2N + 4}} \leq n \Rightarrow w = 1## which value we plug into the function being maximized to determine the maximum to be ##\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N + 2}}}  \leq \sum\limits_{k = 1}^n {{w^{2N + 2}}}  = n## thus ##S_{N + 2}^n \subset S_{N + 1}^n##. Hence the lemma is true.

#### Theorem 3.2: Equivalence of ##S^n## to ##P^n##

If ##S_N^n: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N}}}  \leq n} \right\}## and if ##{P^n}: = \left\{ {\left( {{x_1},{x_2}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}|\left| {{x_k}} \right| \leq {b_k}\forall k \leq n} \right\}##. Then ##{S^n} \subset  \cdots  \subset S_{N + 1}^n \subset S_N^n \subset  \cdots  \subset S_1^n## where ##{S^n}: = \bigcap\limits_{j = 1}^\infty  {S_j^n} ## and ##{S^n} = {P^n}##.

##### Proof:

That ##{S^n} \subset  \cdots  \subset S_{N + 1}^n \subset S_N^n \subset  \cdots  \subset S_1^n## we have lemma 3.1 and that if ##\vec x \in {S^n}: = \bigcap\limits_{j = 1}^\infty  {S_j^n} ##, then ##\vec x \in S_N^n\forall N \in {\mathbb{Z}^ + }##. For the equality ##{S^n} = {P^n}##, suppose ##\exists \vec y \in S\left( {{S^n},{P^n}} \right) = \left( {{S^n} – {P^n}} \right) \cup \left( {{P^n} – {S^n}} \right)## , where ##S\left( {A,B} \right)## is the symmetric difference of sets A and B. Then either

(i) $$\begin{gathered}\exists \vec y \in \left( {{S^n} – {P^n}} \right) = \left\{ {\vec x \in {\mathbb{R}^n}|\left( {\vec x \in {S^n}} \right) \wedge \left( {\vec x \notin {P^n}} \right)} \right\} \\ = \left\{ {\vec x \in {\mathbb{R}^n}|\vec x \in {S^n} = \mathop {\lim }\limits_{M \to \infty } \,\bigcap\limits_{j = 1}^M {S_j^n} = \mathop {\lim }\limits_{M \to \infty } S_M^n\,} \right\} \cap \left\{ {\vec x \in {\mathbb{R}^n}|\exists k \in {\mathbb{Z}^ + } \mathrel\backepsilon \left( {1 \leq k \leq n} \right) \wedge \left( {\tfrac{{\left| {{x_k}} \right|}}{{{b_k}}} > 1} \right)} \right\} \\ \end{gathered}$$

where we have applied $${S^n} \subset \cdots \subset S_{N + 1}^n \subset S_N^n \subset \cdots \subset S_1^n$$ in the evaluation of the limit. Furthermore $$\exists k \in {\mathbb{Z}^ + } \mathrel\backepsilon \left( {1 \leq k \leq n} \right) \wedge \left( {\tfrac{{\left| {{x_k}} \right|}}{{{b_k}}} > 1} \right) \Rightarrow \exists \varepsilon > 0 \mathrel\backepsilon \tfrac{{\left| {{x_k}} \right|}}{{{b_k}}} = 1 + \varepsilon \Rightarrow \exists M \in {\mathbb{Z}^ + } \mathrel\backepsilon N \geq M \Rightarrow \sum\limits_{j = 1}^n {{{\left( {\frac{{{x_j}}}{{{b_j}}}} \right)}^{2N}}} > n$$

Even for the minimal condition that ##{x_j} = 0\forall j \ne k \leq n## we have by the binomial theorem that $$\sum\limits_{j = 1}^n {{{\left( {\frac{{{x_j}}}{{{b_j}}}} \right)}^{2N}}} = {\left( {1 + \varepsilon } \right)^{2N}} = \sum\limits_{j = 1}^{2N} {\left( {\left. {\begin{array}{*{20}{c}}{2N} \\ j \end{array}} \right)} \right.} \,{\varepsilon ^j} = 1 + 2N\varepsilon + N\left( {2N – 1} \right){\varepsilon ^2} + \cdots$$  so choose ##N \in {\mathbb{Z}^ + } \mathrel\backepsilon  2N\varepsilon  \geq n \Rightarrow N = \left\lceil {\tfrac{{n – 1}}{{2\varepsilon }}} \right\rceil##   and hence ##\left( {{S^n} – {P^n}} \right) = \emptyset##  a contradiction.

Or (ii)

$$\begin{gathered}\exists \vec y \in \left( {{P^n} – {S^n}} \right) = \left\{ {\vec x \in {\mathbb{R}^n}|\left( {\vec x \in {P^n}} \right) \wedge \left( {\vec x \notin {S^n}} \right)} \right\} \\ = \left\{ {\vec x \in {\mathbb{R}^n}|\tfrac{{\left| {{x_k}} \right|}}{{{b_k}}} \leq 1{\text{ for }}k = 1,2, \ldots ,n\,} \right\} \cap \left\{ {\vec x \in {\mathbb{R}^n}|\vec x \notin \bigcap\limits_{j = 1}^\infty {S_j^n} } \right\} \\ \end{gathered}$$

we have from the left-hand set ##\tfrac{{\left| {{x_k}} \right|}}{{{b_k}}} \leq 1## for ##k = 1,2, \ldots ,n##, which implies that $$\forall N \in {\mathbb{Z}^ + },\sum\limits_{j = 1}^n {{{\left( {\frac{{{x_j}}}{{{b_j}}}} \right)}^{2N}}} \leq n \cdot {1^{2N}} = n \Rightarrow \vec x \in \bigcap\limits_{j = 1}^\infty {S_j^n}$$ so that ##\left( {{P^n} – {S^n}} \right) = \emptyset ##, a contradiction. Having found contradictions in both cases (i) and (ii) we conclude $$S\left( {{S^n},{P^n}} \right) = \left( {{S^n} – {P^n}} \right) \cup \left( {{P^n} – {S^n}} \right) = \emptyset$$ a contradiction to the former assumption proving ##{S^n} = {P^n}##.

### Content of Hyperellipsoid

The hypervolume described by in the first orthant (or hyperoctant) by $${E^n}: = \left\{ {\vec x \in {\mathbb{R}^n}|{x_i} \geq 0\forall i,\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{a_k}}}} \right)}^2} \leq 1} } \right\}$$ is the generalization of the ellipsoid to ##{\mathbb{R}^n}##, the hyperellipsoid with semi-axes ##\left\{ {{a_i}} \right\},i = 1,2, \ldots ,n##. The content of this hyperellipsoid is given then by corollary 2.4 which gives here

##\begin{gathered}{\text{content}}\left( {{\text{hyperellipsoid}}} \right) = {2^n}\iint {\mathop  \cdots \limits_{{E^n}} \int {d{z_n} \ldots d{z_2}d{z_1}} } \\ = {2^n}{{\prod\limits_{q = 1}^n {\left[ {\frac{{{a_p}}}{2}\Gamma \left( {\frac{1}{2}} \right)} \right]} } \mathord{\left/ {\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{{a_p}}}{2}\Gamma \left( {\frac{1}{2}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{2}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{2}} } \right)}}  \\ = \tfrac{{{\pi ^{\tfrac{n}{2}}}\prod\limits_{q = 1}^n {{a_p}} }}{{\Gamma \left( {1 + \tfrac{n}{2}} \right)}} \\ \end{gathered}##

## Unit Hypercube Integrals

For reference, we state here the Lebesgue Dominated Convergence Theorem,

#### Theorem 4.1 (DCT): Lebesgue Dominated Convergence Theorem

Suppose ##E \in \mathfrak{M}## (the family of measurable sets). Let ##\left\{ {{f_n}} \right\}##  be a sequence of measurable functions such that ##{f_n}\left( x \right) \to f\left( x \right)## almost everywhere on E as ##n \to \infty ##. If there exists a function ##g \in \mathfrak{L}\left( \mu  \right)## on E (g is Lebesgue integrable with respect to ##\mu ## on E), such that ##\left| {{f_n}\left( x \right)\,} \right| \leq g\left( x \right)##, for ##n = 1,2,3, \ldots## and ##\forall x \in E##, then $$\mathop {\lim }\limits_{n \to \infty } \,\int_E {{f_n}d\mu } = \int_E {fd\mu }$$[2]

#### Theorem 4.2: Dominated Convergence Theorem with Nesting Property

Let ##{A_n},A \subset {\mathbb{R}^n}## and ##{A_n},A \in \mathfrak{M}## for ##n \in {\mathbb{Z}^ + }## such that $$A \subset \cdots \subset {A_{n + 1}} \subset {A_n} \subset \cdots \subset {A_1}$$ and let ##A: = \bigcap\limits_{j = 1}^\infty  {{A_j}} ##. Then for Lebesgue measurable ##f:{\mathbb{R}^n} \to \mathbb{C}## define the set function ##\phi :{A_1} \to \mathbb{C}## by

$$\phi \left( E \right): = \int_E {fd\mu } ,\forall E \subset {A_1}$$ Then $$\mathop {\lim }\limits_{n \to \infty } \,\phi \left( {{A_n}} \right) = \phi \left( A \right)$$ which is to say explicitly that $$\mathop {\lim }\limits_{n \to \infty } \,\int_{{A_n}} {fd\mu } = \int_A {fd\mu }$$

##### Proof:

Define ##{\chi _E}\left( x \right): = \left\{ {\begin{array}{*{20}{c}}{1,}&{x \in E} \\ {0,}&{x \notin E}\end{array}} \right.##

the characteristic function of the set E. Then let ##{f_n}: = f \circ {\chi _{\bigcap\limits_{j = 1}^n {{A_j}} }} = \left\{ {\begin{array}{*{20}{c}}{f\left( x \right),}&{x \in {A_n}} \\ {0,}&{x \notin {A_n}}\end{array}} \right.##

where the nesting property has been used. Also let ##f: = f \circ {\chi _A} = \left\{ {\begin{array}{*{20}{c}}{f\left( x \right),}&{x \in A} \\ {0,}&{x \notin A}\end{array}} \right.##

Then $${f_1}\left( x \right) \geq {f_2}\left( x \right) \geq \cdots \geq {f_n}\left( x \right) \geq {f_{n + 1}}\left( x \right) \geq \cdots \geq f\left( x \right)$$

hence by the Lebesgue Dominated Convergence Theorem with $${f_1}\left( x \right) \geq \left| {{f_n}\left( x \right)} \right|\,,\forall x \in {\mathbb{R}^n},\forall n \in {\mathbb{Z}^ + }$$

we have

$$\mathop {\lim }\limits_{n \to \infty } \,\phi \left( {{A_n}} \right) = \mathop {\lim }\limits_{n \to \infty } \,\int_{{A_n}} {fd\mu } = \int_A {fd\mu } = \phi \left( A \right)$$

the required result.

#### Example 4.3: Content of the Orthotope

As an example of Theorem 4.2 consider the orthotope $${S^{n\left( + \right)}}: = \bigcap\limits_{j = 1}^\infty {S_j^{n\left( + \right)}}$$ where we define $$S_N^{n\left( + \right)}: = \left\{ {\vec x \in {\mathbb{R}^n}|{x_i} \geq 0\forall i,\sum\limits_{k = 1}^n {{{\left( {\frac{{{x_k}}}{{{b_k}}}} \right)}^{2N}}} \leq n} \right\}$$

Note that by lemma 3.1 ##S_N^{n\left(  +  \right)}## and ##{S^{n\left(  +  \right)}}## satisfy the nesting property in the hypotheses of theorem 4.2 so that we can evaluate the content integral by corollary 2.2,

$$\begin{gathered}{\text{content}}\left( {{S^{n\left( + \right)}}} \right) = \mathop {\lim }\limits_{N \to \infty } \,\iint {\mathop \cdots \limits_{S_N^{n\left( + \right)}} \int {d{z_n} \ldots d{z_2}d{z_1}} } \\ = \mathop {\lim }\limits_{N \to \infty } \,\left\{ {{{{n^{\sum\limits_{k = 1}^n {\tfrac{1}{{2N}}} }}\prod\limits_{q = 1}^n {\left[ {\frac{{{b_p}}}{{2N}}\Gamma \left( {\frac{1}{{2N}}} \right)} \right]} } \mathord{\left/{\vphantom {{{n^{\sum\limits_{k = 1}^n {\tfrac{1}{{2N}}} }}\prod\limits_{q = 1}^n {\left[ {\frac{{{b_p}}}{{2N}}\Gamma \left( {\frac{1}{{2N}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{2N}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{2N}}} } \right)}}} \right\} \\ = \mathop {\lim }\limits_{N \to \infty } \,\left[ {{n^{\tfrac{n}{{2N}}}}\tfrac{{{\Gamma ^n}\left( {\frac{1}{{2N}}} \right)\prod\limits_{q = 1}^n {{b_p}} }}{{{{\left( {2N} \right)}^n}\Gamma \left( {1 + \tfrac{n}{{2N}}} \right)}}} \right] = \prod\limits_{q = 1}^n {\left( {{b_p}} \right) \cdot } \underbrace {\mathop {\lim }\limits_{N \to \infty } \,\left( {{n^{\tfrac{n}{{2N}}}}} \right)}_{ = {n^0}} \cdot \underbrace {\mathop {\lim }\limits_{N \to \infty } \left[ {\tfrac{{{\Gamma ^n}\left( {1 + \frac{1}{{2N}}} \right)}}{{\Gamma \left( {1 + \tfrac{n}{{2N}}} \right)}}} \right]}_{ = 1{\text{ by exercise 1.3}}} = \prod\limits_{q = 1}^n {{b_p}} \\ \end{gathered}$$

as it must.

#### Example 4.4: The Zeta Function

Show that $$Z\left( n \right): = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{k = 1}^n {{x_k}} } \right)}^{ – 1}}d{x_n} \ldots d{x_2}d{x_1}} } } = \zeta \left( n \right)\forall n \in {\mathbb{Z}^ + } – \left\{ 1 \right\}$$

where ##\zeta \left( {\, \cdot \,} \right)## is the Riemann zeta function. The domain of integration is a unit hypercube with one vertex at the origin and the vertex which is farthest from said point is at ##\left( {1,1, \ldots ,1} \right)## suggesting the use of the sequence of sets here defined throughout the remnant of this Insight $$C_N^n: = \left\{ {\vec x \in {\mathbb{R}^n}|{x_i} \geq 0\forall i,\sum\limits_{k = 1}^n {x_k^{2N}} \leq n – 1} \right\}$$ and also define the set ##{C^n}: = \bigcap\limits_{j = 1}^\infty  {C_j^n} ##. Then we have

$$\begin{gathered}Z\left( n \right) = \mathop {\lim }\limits_{N \to \infty } \,\int {\int_{C_N^n} { \cdots \int {\tfrac{{\prod\nolimits_{i = 1}^n {d{x_i}} }}{{1 – \prod\nolimits_{k = 1}^n {{x_k}} }}} } } = \mathop {\lim }\limits_{N \to \infty } \,\int {\int_{C_N^n} { \cdots \int {\sum\limits_{k = 1}^\infty {\prod\limits_{i = 1}^n {\left( {x_i^{k – 1}d{x_i}} \right)} } } } } \\ = \mathop {\lim }\limits_{N \to \infty } \,\sum\limits_{k = 1}^\infty {\int {\int_{C_N^n} { \cdots \int {\prod\limits_{i = 1}^n {\left( {x_i^{k – 1}d{x_i}} \right)} } } } } \\ = \mathop {\lim }\limits_{N \to \infty } \,\sum\limits_{k = 1}^\infty {{{\left( {n – 1} \right)}^{\sum\limits_{p = 1}^n {\frac{k}{{2N}}} }}{{\prod\limits_{q = 1}^n {\left[ {\frac{{{1^k}}}{{2N}}\Gamma \left( {\frac{k}{{2N}}} \right)} \right]} } \mathord{\left/ {\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{{1^k}}}{{2N}}\Gamma \left( {\frac{k}{{2N}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{i = 1}^n {\frac{k}{{2N}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{i = 1}^n {\frac{k}{{2N}}} } \right)}}} \\ = \mathop {\lim }\limits_{N \to \infty } \,\left\{ {\sum\limits_{k = 1}^\infty {\left[ {{{\left( {n – 1} \right)}^{\tfrac{{nk}}{{2N}}}} \cdot \tfrac{1}{{{k^n}}} \cdot \tfrac{{{\Gamma ^n}\left( {1 + \tfrac{k}{{2N}}} \right)}}{{\Gamma \left( {1 + \tfrac{{nk}}{{2N}}} \right)}}} \right]} } \right\} \\ = \sum\limits_{k = 1}^\infty {\tfrac{1}{{{k^n}}} \cdot \underbrace {\mathop {\lim }\limits_{N \to \infty } \,{{\left( {n – 1} \right)}^{\tfrac{{nk}}{{2N}}}}}_{ = {{\left( {n – 1} \right)}^0}} \cdot \underbrace {\mathop {\lim }\limits_{N \to \infty } \,\tfrac{{{\Gamma ^n}\left( {1 + \tfrac{k}{{2N}}} \right)}}{{\Gamma \left( {1 + \tfrac{{nk}}{{2N}}} \right)}}}_{ = 1{\text{ by exercise 1.3}}}} = \sum\limits_{k = 1}^\infty {\tfrac{1}{{{k^n}}}} = \zeta \left( n \right) \\ \end{gathered}$$

This was a well-known integral, but likely demonstrated a new way to achieve a known result. We shall use this technique of evaluating improper multiple integrals over the unit hypercube on many functions in this section. But first,

#### An alternate evaluation of the above Zeta function example:

$$\begin{gathered}Z\left( n \right) = \int {\int_{{C^n}} { \cdots \int {\tfrac{{\prod\nolimits_{i = 1}^n {d{x_i}} }}{{1 – \prod\nolimits_{k = 1}^n {{x_k}} }}} } } = \int {\int_{{C^{n – 1}}} { \cdots \int {\ln \left( {\tfrac{1}{{1 – \prod\nolimits_{k = 1}^{n – 1} {{x_k}} }}} \right) \cdot \prod\limits_{i = 1}^{n – 1} {\tfrac{{d{x_i}}}{{{x_i}}}} } } } \\ = \int {\int_{{C^{n – 1}}} { \cdots \int {\ln \left[ {\prod\limits_{q = 0}^\infty {\left( {1 + \prod\limits_{k = 1}^{n – 1} {x_k^{{2^q}}} } \right)} } \right] \cdot \prod\limits_{i = 1}^{n – 1} {\tfrac{{d{x_i}}}{{{x_i}}}} } } } \\ = \sum\limits_{q = 0}^\infty {\int {\int_{{C^{n – 1}}} { \cdots \int {\ln \left( {1 + \prod\limits_{k = 1}^{n – 1} {x_k^{{2^q}}} } \right) \cdot \prod\limits_{i = 1}^{n – 1} {\tfrac{{d{x_i}}}{{{x_i}}}} } } } } = \mathop {\lim }\limits_{N \to \infty } \,\sum\limits_{q = 0}^\infty {\sum\limits_{k = 1}^\infty {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{k}\int {\int_{C_N^{n – 1}} { \cdots \int {\prod\limits_{k = 1}^{n – 1} {x_k^{k{2^q} – 1}d{x_k}} } } } } } \\ = \mathop {\lim }\limits_{N \to \infty } \,\sum\limits_{q = 0}^\infty {\sum\limits_{k = 1}^\infty {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{k}{{\left( {n – 2} \right)}^{\sum\limits_{p = 1}^n {\frac{{k{2^q}}}{{2N}}} }}{{\prod\limits_{i = 1}^{n – 1} {\left[ {\frac{1}{{2N}}\Gamma \left( {\frac{{k{2^q}}}{{2N}}} \right)} \right]} } \mathord{\left/ {\vphantom {{\prod\limits_{i = 1}^{n – 1} {\left[ {\frac{1}{{2N}}\Gamma \left( {\frac{{k{2^q}}}{{2N}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{j = 1}^{n – 1} {\frac{{k{2^q}}}{{2N}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{j = 1}^{n – 1} {\frac{{k{2^q}}}{{2N}}} } \right)}}} } \\ = \sum\limits_{q = 0}^\infty {\sum\limits_{k = 1}^\infty {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{{{k^n}{2^{\left( {n – 1} \right)q}}}} \cdot \underbrace {\mathop {\lim }\limits_{N \to \infty } \,{{\left( {n – 2} \right)}^{\frac{{\left( {n – 1} \right)k{2^q}}}{{2N}}}}}_{ = {{\left( {n – 2} \right)}^0}} \cdot \underbrace {\mathop {\lim }\limits_{N \to \infty } \,\tfrac{{{\Gamma ^{n – 1}}\left( {1 + \frac{{k{2^q}}}{{2N}}} \right)}}{{\Gamma \left( {1 + \tfrac{{\left( {n – 1} \right)k{2^q}}}{{2N}}} \right)}}}_{ = 1{\text{ by excercise 1.3}}}} } \\ = \sum\limits_{q = 0}^\infty {\sum\limits_{k = 1}^\infty {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{{{k^n}}} \cdot {{\left( {\tfrac{1}{{{2^{n – 1}}}}} \right)}^q}} } = \sum\limits_{k = 1}^\infty {\left[ {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{{{k^n}}}\sum\limits_{q = 0}^\infty {{{\left( {\tfrac{1}{{{2^{n – 1}}}}} \right)}^q}} } \right]} \\ = {\left( {1 – {2^{1 – n}}} \right)^{ – 1}}\sum\limits_{k = 1}^\infty {\tfrac{{{{\left( { – 1} \right)}^{k – 1}}}}{{{k^n}}}} = \sum\limits_{k = 1}^\infty {\tfrac{1}{{{k^n}}}} = \zeta \left( n \right) \\ \end{gathered}$$

The reader may use this technique of evaluating multiple integrals over the unit hypercube to verify the following

#### Summary of Functions Represented as Multiple Integrals Over the Unit Hypercube:

The Lerch Transcendent: ##\Phi \left( {z,n,y} \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^q}}}{{{{\left( {q + y} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{y – 1}d{\lambda _k}} } } } }##

Legendre Chi Function: ##{\chi _n}\left( z \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^{2q + 1}}}}{{{{\left( {2q + 1} \right)}^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – {z^2}\prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Polygamma Function: ##{\psi _n}\left( z \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^{n + 1}}n!}}{{{{\left( {z + q} \right)}^{n + 1}}}} = {{\left( { – 1} \right)}^{n + 1}}n!\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^{n + 1} {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^{n + 1} {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Polylogarithm of Order n: ##{\text{L}}{{\text{i}}_n}\left( z \right): = \sum\limits_{q = 1}^\infty  {\frac{{{z^q}}}{{{q^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hurwitz Zeta Function: ##\zeta \left( {n,z} \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {q + z} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Riemann Zeta Function: ##\zeta \left( n \right): = \sum\limits_{q = 1}^\infty  {\frac{1}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Dirichlet Beta Function: ##\beta \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^q}}}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Dirichlet Eta Function: ##\eta \left( n \right): = \sum\limits_{q = 1}^\infty  {\frac{{{{\left( { – 1} \right)}^{q – 1}}}}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Dirichlet Lambda Function: ##\lambda \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Exercises 4

Answers Ex. – Section 4 – Unit Hypercube Integrals

4.1) Verify the remaining unit hypercube integral identities from the above summary.

## Fractional Integrals

We will delve into fractional integrals in this section, motivated by the restriction to positive integers n imposed on our n-dimensional integral formulas for the special functions in the last section, we here explore the connection to fractional integral operators, specifically the left Hadamard fractional integral operator ##_aI_x^\alpha ## , defined as

#### Definition 5.1: left Hadamard fractional integral operator

For ##0 < a < x < \infty ,\Re \left[ \alpha  \right] > 0,\, _a^HI_x^\alpha f\left( x \right) = \tfrac{1}{{\Gamma \left( \alpha  \right)}}\int_{t = a}^x {{{\log }^{\alpha  – 1}}\left( {\tfrac{x}{t}} \right)f\left( t \right)\tfrac{{dt}}{t}} ##

assuming the integral is convergent. Would it surprise you to know we’ve already met a fractional integral of this type? Back in the proof of theorem 1.2, upon application of L’Hospital’s rule we saw that $$\Gamma \left( z \right) = \int_0^1 {{{\ln }^{z – 1}}\left( {\frac{1}{y}} \right)dy} \Rightarrow \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^HI_x^z\left( 1 \right) = 1$$

We recognize that when ##a##, and ##x## are chosen such that for the range of values of t dictated by the bounds of integration ##\log \left( {\tfrac{x}{t}} \right) < 0##, in general ##{\log ^{\alpha  – 1}}\left( {\tfrac{x}{t}} \right)## is multi-valued, so that we may have instead a single-valued function provided that ##\log \left( {\tfrac{x}{a}} \right) \geq 0##  thereby defined we introduce the modified left Hadamard fractional operator,

#### Definition 5.2: Modified Left Hadamard Fractional Operator

For $$0 < a < x < \infty ,\Re \left[ \alpha \right] > 0,\,_a^{eH}I_x^\alpha f\left( x \right) = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_{u = 0}^{\log \left( {\tfrac{x}{a}} \right)} {{u^{\alpha – 1}}f\left( {x{e^{ – u}}} \right)du}$$

where we have simply substituted ##u = \log \left( {\tfrac{x}{t}} \right)## in the integral of definition 5.1 and ##{u^{\alpha  – 1}}## is taken as it’s principle value.

The left Hadamard fractional integral operator interpolates the n-fold iterated integral

#### Equation 5.3: Interpolation of the n-fold Integral by the Hadamard Fractional Integral Operator

$$\int_a^x {\int_a^{{x_1}} { \cdots \int_a^{{x_{n – 1}}} {f\left( {{x_n}} \right)\tfrac{{d{x_n} \ldots d{x_1}}}{{{x_n} \cdots {x_1}}}} } } = \,_a^HI_x^nf\left( x \right) = \tfrac{1}{{\left( {n – 1} \right)!}}\int_a^x {{{\log }^{n – 1}}\left( {\tfrac{x}{t}} \right)f\left( t \right)dt}$$

##### Proof:

The proof is by induction on n:

(i) base case of ##n = 1## is obvious.

(ii) Assume that equation 5.3 holds for some fixed positive integer n. Then,

$$\begin{gathered}\int_a^x {\int_a^{{x_1}} { \cdots \int_a^{{x_n}} {f\left( {{x_{n + 1}}} \right)\tfrac{{d{x_{n + 1}} \ldots d{x_1}}}{{{x_{n + 1}} \cdots {x_1}}}} } } = \int_a^x {\left[ {\int_a^{{x_1}} { \cdots \int_a^{{x_n}} {f\left( {{x_{n + 1}}} \right)\tfrac{{d{x_{n + 1}} \ldots d{x_2}}}{{{x_{n + 1}} \cdots {x_2}}}} } } \right]\tfrac{{d{x_1}}}{{{x_1}}}} \\ = \int_a^x {\left[ {\tfrac{1}{{\left( {n – 1} \right)!}}\int_a^{{x_1}} {{{\log }^{n – 1}}\left( {\tfrac{{{x_1}}}{t}} \right)f\left( t \right)\tfrac{{dt}}{t}} } \right]\tfrac{{d{x_1}}}{{{x_1}}}} \\ = \tfrac{1}{{\left( {n – 1} \right)!}}\int_a^x {\int_t^x {{{\log }^{n – 1}}\left( {\tfrac{{{x_1}}}{t}} \right)f\left( t \right)\tfrac{{d{x_1}dt}}{{{x_1}t}}} } \\ = \tfrac{1}{{\left( {n – 1} \right)!}}\sum\limits_{k = 0}^{n – 1} {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}{n – 1} \\ k \end{array}} \right)\int_a^x {\int_t^x {{{\log }^k}\left( {{x_1}} \right){{\log }^{n – k – 1}}\left( t \right)f\left( t \right)\tfrac{{d{x_1}dt}}{{{x_1}t}}} } } \\ = \tfrac{1}{{\left( {n – 1} \right)!}}\sum\limits_{k = 0}^{n – 1} {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}{n – 1} \\ k \end{array}} \right)\int_a^x {{{\log }^{n – k – 1}}\left( t \right)f\left( t \right)\tfrac{1}{t}\left[ {\int_t^x {{{\log }^k}\left( {{x_1}} \right)\tfrac{{d{x_1}}}{{{x_1}}}} } \right]dt} } \\ = \tfrac{1}{{\left( {n – 1} \right)!}}\sum\limits_{k = 0}^{n – 1} {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}{n – 1} \\ k \end{array}} \right)\tfrac{1}{{k + 1}}\int_a^x {{{\log }^{n – k – 1}}\left( t \right)f\left( t \right)\tfrac{1}{t}\left( {{{\log }^{k + 1}}x – {{\log }^{k + 1}}t} \right)dt} } \\ = \tfrac{1}{{n!}}\sum\limits_{k = 0}^n {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right)\int_a^x {{{\log }^{n – k}}\left( t \right){{\log }^k}\left( x \right)f\left( t \right)\tfrac{{dt}}{t}} = } \tfrac{1}{{n!}}\int_a^x {{{\log }^n}\left( {\tfrac{x}{t}} \right)f\left( t \right)\tfrac{{dt}}{t}} \\ \end{gathered}$$

and the proof is complete.

#### Example 5.4: An Integral for the Riemann Zeta Function

We will use the change of variables[3] ##{y_k} = \prod\limits_{i = 1}^k {{\lambda _i}} ,k = 1,2, \ldots ,n## on the integral derived in the previous section for the Riemann zeta function to formulate an integral that represents the function for complex values of the argument via equation 5.3. Said integral, namely

$$\zeta \left( n \right): = \sum\limits_{q = 1}^\infty {\frac{1}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }$$

Note that the for given change of variables we have ##{\lambda _1} = {y_1},{\lambda _k} = \tfrac{y_k}{y_{k – 1}},k = 2,3, \ldots ,n##, hence $$\tfrac{{\partial {\lambda _i}}}{{\partial {y_j}}} = \left\{ {\begin{array}{*{20}{c}}{1,}&{i = j = 1} \\ {\tfrac{1}{{{y_{i – 1}}}},}&{i = j \ne 1} \\ { – \tfrac{{{y_i}}}{{y_{i – 1}^2}},}&{i = j – 1} \\ {0,}&{{\text{otherwise}}} \end{array}} \right.$$ whence the Jacobian determinant is the product along the diagonal,

##\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}## and we notice that this change of variables maps the unit hypercube ##{\left[ {0,1} \right]^n}## to the simplex ##\left\{ {\vec y \in {\mathbb{R}^n}|0 \leq {y_1} \leq 1,0 \leq {y_i} \leq {y_{i – 1}},i = 2,3, \ldots ,n} \right\}## so we replace the upper bound of ##{y_1}## with x to get

$$\begin{gathered} \zeta \left( n \right) = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } \\ = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \int_a^x {\int_a^{{y_1}} { \cdots \int_a^{{y_{n – 1}}} {\tfrac{{{y_n}}}{{1 – {y_n}}} \cdot \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_n} \ldots {y_1}}}} } } \\ = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^n\left( {\tfrac{x}{{1 – x}}} \right) \\ = \tfrac{1}{{\left( {n – 1} \right)!}}\int_0^\infty {{u^{n – 1}}\tfrac{{{e^{ – u}}}}{{1 – {e^{ – u}}}}du} \\ \end{gathered}$$

This suggests we should investigate whether ##\zeta \left( \alpha  \right)\mathop  = \limits^? \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{{1 – x}}} \right). ## We have

$$\begin{gathered}\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{{1 – x}}} \right) = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty {\tfrac{{{u^{\alpha – 1}}{e^{ – u}}}}{{1 – {e^{ – u}}}}du} = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty {{u^{\alpha – 1}}{e^{ – u}}\sum\limits_{k = 0}^\infty {{e^{ – ku}}} du} \\ = \tfrac{1}{{\Gamma \left( \alpha \right)}}\sum\limits_{k = 0}^\infty {\int_0^\infty {{u^{\alpha – 1}}{e^{ – \left( {k + 1} \right)u}}du} } \\ = \tfrac{1}{{\Gamma \left( \alpha \right)}}\sum\limits_{k = 0}^\infty {\tfrac{1}{{{{\left( {k + 1} \right)}^\alpha }}}\int_0^\infty {{t^{\alpha – 1}}{e^{ – t}}dt} } = \sum\limits_{k = 1}^\infty {\tfrac{1}{{{k^\alpha }}}} = \zeta \left( \alpha \right),\Re \left[ \alpha \right] > 1 \\ \end{gathered}$$

as hoped for. End example.

A similar investigation suggests that the Lerch Transcendent $$\Phi \left( {z,\alpha ,y} \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^n\left( {\tfrac{{{x^y}}}{{1 – zx}}} \right),\Re \left[ \alpha \right] > 1$$ This is equivalent to $$\tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty {\tfrac{{{u^{\alpha – 1}}{e^{ – yu}}}}{{1 – z{e^{ – u}}}}du} = \sum\limits_{q = 0}^\infty {\frac{{{z^q}}}{{{{\left( {q + y} \right)}^\alpha }}}}$$ and the evaluation is very similar to the zeta function example above, as such the proof of this is left as an exercise to the reader.

## Exercises 5 – Fractional Integrals

5.1) Prove that ##\tfrac{1}{{\Gamma \left( \alpha  \right)}}\int_0^\infty  {\tfrac{{{u^{\alpha  – 1}}{e^{ – yu}}}}{{1 – z{e^{ – u}}}}du}  = \Phi \left( {z,\alpha ,y} \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^q}}}{{{{\left( {q + y} \right)}^\alpha }}}} ,\Re \left[ \alpha  \right] > 1##.

Removed exercise 5.2) Use the result of exercise 5.1 to prove that ##\Phi \left( {z,\alpha ,\tfrac{y}{2}} \right) – \Phi \left( { – z,\alpha ,\tfrac{y}{2}} \right) = {2^{1 – \alpha }}\Phi \left( {{z^2},\alpha ,y} \right)##.

Removed exercise 5.3) Use the result of exercise 5.1 to prove that ##\forall n \in {\mathbb{Z}^*}: = {\mathbb{Z}^ + } \cup \left\{ 0 \right\}, \Phi \left( {z,\alpha ,y + n} \right) =  – \sum\limits_{k = 0}^n {\frac{{{z^k}}}{{{{\left( {y + k} \right)}^\alpha }}}}  + {z^n}\Phi \left( {z,\alpha ,y} \right)##.

5.4) Determine the fractional Integral Analogs (and prove equivalence to the series) of the following special functions represented as integrals (that is follow the Example 5.4 for the given function):

a) Legendre Chi Function: ##{\chi _n}( z ) : = \sum_{q = 0}^\infty  \frac{z^{2q + 1}}{( 2q + 1 )^n} = z\int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – z^2\prod_{q = 1}^n \lambda _q^2  \right) ^{ – 1}\prod_{k = 1}^n d{\lambda _k}##

Hint: $$\chi_\alpha (z)=z\begin{gathered}\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{1 – x^2 z^2}} \right) = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – z^2 e^{ -2 u}}du \end{gathered}$$

b) Polygamma Function: ##{\psi _n}\left( z \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^{n + 1}}n!}}{{{{\left( {z + q} \right)}^{n + 1}}}} = {{\left( { – 1} \right)}^{n + 1}}n!\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^{n + 1} {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^{n + 1} {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\psi _{\alpha}\left( z \right) = (-1)^{\alpha +1}\Gamma (\alpha +1)\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x^z}{1-x}} \right) =(-1)^{\alpha +1} \alpha \int_0^\infty u^{\alpha – 1}\tfrac{e^{-uz}}{1 – e^{ – u}}du \end{gathered}$$

c) Polylogarithm of Order n: ##{\text{L}}{{\text{i}}_n}\left( z \right): = \sum\limits_{q = 1}^\infty  {\frac{{{z^q}}}{{{q^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}{\text{L}}{{\text{i}}_\alpha}\left( z \right) = z\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{1-zx}} \right) =\tfrac{z}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 – ze^{ – u}}du \end{gathered}$$

d) Hurwitz Zeta Function: ##\zeta \left( {n,z} \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {q + z} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\zeta \left( {n,z} \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x^z}{1-x}} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{-zu}}{1 – e^{ – u}}du \end{gathered}$$

e) Dirichlet Beta Function: ##\beta \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^q}}}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\beta\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1+x^2} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -2 u}}du \end{gathered}$$

f) Dirichlet Eta Function: ##\eta \left( n \right): = \sum\limits_{q = 1}^\infty  {\frac{{{{\left( { – 1} \right)}^{q – 1}}}}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\eta\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1+x} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -u}}du \end{gathered}$$

g) Dirichlet Lambda Function: ##\lambda \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\lambda\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1-x^2} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 -e^{ -2 u}}du \end{gathered}$$

### Exercises 4 – Unit Hypercube Integrals

4.1) Verify the remaining unit hypercube integral identities from the above summary.

a) The Lerch Transcendent: ##\Phi \left( {z,n,y} \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^q}}}{{{{\left( {q + y} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{y – 1}d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\Phi ( z,n,y ) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – z\prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^n \lambda _k^{y – 1}d{\lambda _k} \\ &=& \lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n}} \int \sum\limits_{k=0}^\infty z^k\prod\limits_{q = 1}^n \lambda _q^{k+y-1}d{\lambda _q} \\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \Phi \left( {z,n,y} \right) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^k {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+y}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+y}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+y}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+y}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+y}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+y}}}{{{2N}}}} }}}_{=1}\sum\limits_{k=0}^\infty z^k {{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+y}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+y}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(k+y)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(k+y)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^k \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(k+y)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{k+y}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(k+y)^n}\Gamma ^n\left( \tfrac{k+y}{2N} \right) } \Gamma \left( 1 +\tfrac{(k+y)n}{2N} \right) } \right. } }_{ =1 \text{ Limit 3.3} } \\ &=& \sum\limits_{k=0}^\infty \tfrac{z^k}{(k+y)^n} \\ \end{eqnarray*}$$

b) Legendre Chi Function: ##\chi _n ( z ) = z \int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – z^2\prod\limits_{q = 1}^n \lambda _q^2 \right) ^{ – 1} \prod\limits_{k = 1}^n d{\lambda _k} ##

$$\begin{eqnarray*}\chi _n ( z ) &=& z\int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – z^2\prod\limits_{q = 1}^n \lambda _q^2 \right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k} \\ &=& z\lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty z^{2k}\prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} } \\ &=& z\lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty z^{2k}\iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \chi _n \left( z \right) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^{2k+1} {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}}_{=1}\sum\limits_{k=0}^\infty z^{2k+1}{{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^{2k+1} \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(2k+1)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{2k+1}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(2k+1)^n}\Gamma ^n\left( \tfrac{2k+1}{2N} \right) } \Gamma \left( 1 +\tfrac{(2k+1)n}{2N} \right) } \right. } }_{ =1 \text{ Ex. 1.3 } } \\ &=& \sum\limits_{k=0}^\infty \tfrac{z^{2k+1}}{(2k+1)^n} \\ \end{eqnarray*}$$

c) Polygamma Function: ##{\psi _n}\left( z \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^{n + 1}}n!}}{{{{\left( {z + q} \right)}^{n + 1}}}} = {{\left( { – 1} \right)}^{n + 1}}n!\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^{n + 1} {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^{n + 1} {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\psi _n ( z ) &=& (-1)^{n+1}n! \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – \prod\limits_{q = 1}^{n+1} \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^{n+1}\lambda_k^{z-1} d{\lambda _k} \\ &=& (-1)^{n+1}n! \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty\iint {\mathop \cdots \limits_{C_N^{n+1}} \int \prod\limits_{q = 1}^{n+1} \lambda _q^{z+k-1}d{\lambda _q} } \\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \psi _n \left( z \right) &=& (-1)^{n+1}n!\lim\limits_{N\to\infty} \sum\limits_{q=0}^\infty {n^{\sum\limits_{p = 1}^{n+1} {\tfrac{{{z+q}}}{{{2N}}}} }}{{\prod\limits_{k = 1}^{n+1} {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{z+q}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{k = 1}^{n+1} {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{z+q}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{\lambda = 1}^{n+1} {\tfrac{{{z+q}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{\lambda = 1}^n {\tfrac{{{z+q}}}{{{2N}}}} } \right)}} \\ &=& (-1)^{n+1}n!\lim\limits_{N\to\infty} \underbrace{{n^{\sum\limits_{p = 1}^{n+1} {\tfrac{{{z+q}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{q=0}^\infty \tfrac{{(2N)^{n+1}}}{(z+q)^{n+1}}\cdot{{ {{\tfrac{{1}}{{{(2N)^{n+1}}}}\Gamma ^{n+1}\left( {1+\tfrac{{{z+q}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{(2N)^{n+1}}}{(z+q)^{n+1}}\cdot\tfrac{{1}}{{{(2N)^{n+1}}}}\Gamma ^{n+1}\left( {1+\tfrac{{{z+q}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(2k+1)(n+1)}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(z+q)(n+1)}}}{{{2N}}}} } \right)}} \\ &=& (-1)^{n+1}n!\lim\limits_{N\to\infty} \sum\limits_{q=0}^\infty \tfrac{1}{(z+q)^{n+1}} \underbrace{\Gamma ^{n+1}\left( 1 + \tfrac{z+q}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(z+q)^{n+1}}\Gamma ^{n+1}\left( 1+\tfrac{z+q}{2N} \right) } \Gamma \left( 1 +\tfrac{(z+q)(n+1)}{2N} \right) } \right. } }_{ =1 \text{ Ex. 1.3 } } \\ &=& \sum\limits_{q=0}^\infty \tfrac{(-1)^{n+1}n!}{(z+q)^{n+1}} \\ \end{eqnarray*}$$

d) Polylogarithm of Order n: ##{\text{L}}{{\text{i}}_n}\left( z \right): = \sum\limits_{q = 1}^\infty  {\frac{{{z^q}}}{{{q^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

$$\begin{eqnarray*}{\text{L}}{{\text{i}}_n} ( z ) &=& z\int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – z\prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k} \\ &=& z\lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty z^{k}\prod\limits_{q = 1}^n \lambda _q^{k}d{\lambda _q} } \\ &=& \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty z^{k+1}\iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{k}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} {\text{L}}{{\text{i}}_n} ( z ) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^{k+1} {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{k=0}^\infty z^{k+1}{{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+1}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(k+1)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(k+1)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty z^{k+1} \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(k+1)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{k+1}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(k+1)^n}\Gamma ^n\left( \tfrac{k+1}{2N} \right) } \Gamma \left( 1 +\tfrac{(k+1)n}{2N} \right) } \right. } }_{ =1 \text{ Ex. 1.3 } } \\ &=& \sum\limits_{k=1}^\infty \tfrac{z^k}{k^n} \\ \end{eqnarray*}$$

e) Hurwitz Zeta Function: ##\zeta \left( {n,z} \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {q + z} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\zeta \left( {n,z} \right) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – \prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^n \lambda_k^{z-1} d{\lambda _k} \\ &=& \lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty \prod\limits_{q = 1}^n \lambda _q^{z+k-1}d{\lambda _q} } \\ &=& \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty \iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{z+k-1}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \zeta \left( {n,z} \right) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{z+k}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{z+k}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{z+k}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{z+k}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{z+k}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{z+k}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{k=0}^\infty {{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{z+k}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{z+k}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(z+k)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(z+k)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(z+k)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{z+k}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(z+k)^n}\Gamma ^n\left( \tfrac{z+k}{2N} \right) } \Gamma \left( 1 +\tfrac{(z+k)n}{2N} \right) } \right. } }_{ =1 \text{ by Ex. 1.3 } } \\ &=& \sum\limits_{k=1}^\infty \tfrac{1}{(z+k)^n} \\ \end{eqnarray*}$$

f) Dirichlet Beta Function: ##\beta ( n ) : = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^q}}}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\beta ( n ) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 + \prod\limits_{q = 1}^n \lambda _q^2\right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k} \\ &=& \lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty (-1)^k\prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} } \\ &=& \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty (-1)^k \iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \beta ( n ) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty (-1)^k {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{k=0}^\infty (-1)^k {{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty (-1)^k \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(2k+1)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{2k+1}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(2k+1)^n}\Gamma ^n\left( \tfrac{2k+1}{2N} \right) } \Gamma \left( 1 +\tfrac{(2k+1)n}{2N} \right) } \right. } }_{ =1 \text{ by Ex. 1.3 } } \\ &=& \sum\limits_{k=0}^\infty \tfrac{(-1)^k}{(2k+1)^n} \\ \end{eqnarray*}$$

g) Dirichlet Eta Function: ##\eta \left( n \right): = \sum\limits_{q = 1}^\infty  {\frac{{{{\left( { – 1} \right)}^{q – 1}}}}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\eta ( n ) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 + \prod\limits_{q = 1}^n \lambda _q\right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k} \\ &=& \lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty (-1)^k\prod\limits_{q = 1}^n \lambda _q^{k}d{\lambda _q} } \\ &=& \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty (-1)^k \iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{k}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \eta ( n ) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty (-1)^k {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{k+1}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{k=0}^\infty (-1)^k {{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+1}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{k+1}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(k+1)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(k+1)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty (-1)^k \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(k+1)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{k+1}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(k+1)^n}\Gamma ^n\left( \tfrac{k+1}{2N} \right) } \Gamma \left( 1 +\tfrac{(k+1)n}{2N} \right) } \right. } }_{ =1 \text{ by Ex. 1.3 } } \\ &=& \sum\limits_{k=1}^\infty \tfrac{(-1)^{k-1}}{k^n} \\ \end{eqnarray*}$$

h) Dirichlet Lambda Function: ##\lambda \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

$$\begin{eqnarray*}\lambda ( n ) &=& \int_0^1 \int_0^1 \cdots \int_0^1 \left( 1 – \prod\limits_{q = 1}^n \lambda _q^2\right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k} \\ &=& \lim\limits_{N\to\infty}\iint {\mathop \cdots \limits_{C_N^n} \int \sum\limits_{k=0}^\infty \prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} } \\ &=& \lim\limits_{N\to\infty}\sum\limits_{k=0}^\infty \iint {\mathop \cdots \limits_{C_N^n} \int \prod\limits_{q = 1}^n \lambda _q^{2k}d{\lambda _q} }\\ \end{eqnarray*}$$

Then by Corollary 2.2, we have

$$\begin{eqnarray*} \lambda ( n ) &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty {(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\tfrac{{1}}{{{2N}}}\Gamma \left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \underbrace{{(n-1)^{\sum\limits_{p = 1}^n {\tfrac{{{2k+1}}}{{{2N}}}} }}}_{=n^0}\sum\limits_{k=0}^\infty {{ {{\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)}} } \mathord{\left/{\vphantom {{ { {\tfrac{{1}}{{{(2N)^n}}}\Gamma ^n\left( {\tfrac{{{2k+1}}}{{{2N}}}} \right)} } } {\Gamma \left( {1 +{\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}}} \right. } {\Gamma \left( {1 + {\tfrac{{{(2k+1)n}}}{{{2N}}}} } \right)}} \\ &=& \lim\limits_{N\to\infty} \sum\limits_{k=0}^\infty \tfrac{1}{(2N)^n} \cdot\tfrac{(2N)^n}{(2k+1)^n} \underbrace{\Gamma ^n\left( 1+ \tfrac{2k+1}{2N} \right) \mathord{\left/{\vphantom {\tfrac{1}{(2N)^n}\cdot\tfrac{(2N)^n}{(2k+1)^n}\Gamma ^n\left( \tfrac{2k+1}{2N} \right) } \Gamma \left( 1 +\tfrac{(2k+1)n}{2N} \right) } \right. } }_{ =1 \text{ by Ex. 1.3 } } \\ &=& \sum\limits_{k=0}^\infty \tfrac{1}{(2k+1)^n} \\ \end{eqnarray*}$$

### Exercises 5 – Fractional Integrals

5.1) Prove that ##\tfrac{1}{{\Gamma \left( \alpha  \right)}}\int_0^\infty  {\tfrac{{{u^{\alpha  – 1}}{e^{ – yu}}}}{{1 – z{e^{ – u}}}}du}  = \Phi \left( {z,\alpha ,y} \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^q}}}{{{{\left( {q + y} \right)}^\alpha }}}} ,\Re \left[ \alpha  \right] > 1##.

$$\begin{eqnarray*}\tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty {\tfrac{{{u^{\alpha – 1}}{e^{ – yu}}}}{{1 – z{e^{ – u}}}}du} &=& \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty u^{\alpha – 1}{e^{ – yu}}\sum_{k=0}^\infty z^k{e^{ – ku} du} \\ &=&\tfrac{1}{{\Gamma \left( \alpha \right)}}\sum_{k=0}^\infty z^k\int_0^\infty { u^{\alpha – 1} e^{ -(y+ k)u} du} \\ &=& \tfrac{1}{{\Gamma \left( \alpha \right)}}\sum_{k=0}^\infty \tfrac{z^k}{(y+k)^{\alpha}}\int_0^\infty { w^{\alpha – 1} e^{ -w} dw}=\sum_{k=0}^\infty \tfrac{z^k}{(y+k)^{\alpha}}\end{eqnarray*}$$

Removed exercise 5.2) Use the result of exercise 5.1 to prove that ##\Phi \left( {z,\alpha ,\tfrac{y}{2}} \right) – \Phi \left( { – z,\alpha ,\tfrac{y}{2}} \right) = {2^{1 – \alpha }}\Phi \left( {{z^2},\alpha ,y} \right)##.

This exercise was removed because it was calculus and algebra beyond the relatively simple series manipulation required to prove the same identity from the series definition.

Removed exercise 5.3) Use the result of exercise 5.1 to prove that ##\forall n \in {\mathbb{Z}^*}: = {\mathbb{Z}^ + } \cup \left\{ 0 \right\}, \Phi \left( {z,\alpha ,y + n} \right) =  – \sum\limits_{k = 0}^n {\frac{{{z^k}}}{{{{\left( {y + k} \right)}^\alpha }}}}  + {z^n}\Phi \left( {z,\alpha ,y} \right)##.

This exercise was removed because it was calculus and algebra beyond the relatively simple series manipulation required to prove the same identity from the series definition.

5.4) Determine the fractional Integral Analogs (and prove equivalence to the series) of the following special functions represented as integrals (that is follow the Example 5.4 for the given function):

a) Legendre Chi Function: ##{\chi _n}\left( z \right): = \sum\limits_{q = 0}^\infty  {\frac{{{z^{2q + 1}}}}{{{{\left( {2q + 1} \right)}^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – {z^2}\prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\chi_\alpha (z)=z\begin{gathered}\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{1 – x^2 z^2}} \right) = \tfrac{1}{{\Gamma \left( \alpha \right)}}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – z^2 e^{ -2 u}}du \end{gathered}$$

First we establish the hint, from

##{\chi _n}\left( z \right) = z\int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – z^2\prod_{q = 1}^n \lambda _q^2 \right) ^{ – 1}\prod_{k = 1}^n d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}\chi _n ( z ) = z{\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n}{ 1 – z^2y_{n}^2} \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = z{\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1 – x^2 z^2} \right) \\ = \tfrac{z}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – z^2 e^{ -2 u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}{\chi _n}( z ) =\tfrac{z}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – z^2 e^{ -2 u}}du\\ = \tfrac{z}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}e^{ – u}\sum_{k=0}^{\infty} z^{2k} e^{ -2k u}du \\ = \tfrac{1}{\Gamma ( \alpha )}\sum_{k=0}^{\infty}z^{2k+1}\int_0^\infty u^{\alpha – 1} e^{ -(2k+1)u}du \\ = \tfrac{1}{\Gamma ( \alpha )}\sum_{k=0}^{\infty}\tfrac{z^{2k+1}}{(2k+1)^{\alpha }}\int_0^\infty \omega ^{\alpha – 1} e^{ -\omega}d{\omega} \\ =\sum_{k=0}^{\infty}\tfrac{z^{2k+1}}{(2k+1)^{\alpha }} \\ \end{gathered}$$

b) Polygamma Function: ##{\psi _n}( z ): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^{n + 1}}n!}}{{{{\left( {z + q} \right)}^{n + 1}}}} = {{\left( { – 1} \right)}^{n + 1}}n!\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod_{q = 1}^{n + 1} {{\lambda _q}} } \right)}^{ – 1}}\prod_{k = 1}^{n + 1} {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\psi _{\alpha}\left( z \right) = (-1)^{\alpha +1}\Gamma (\alpha +1)\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x^z}{1-x}} \right) =(-1)^{\alpha +1} \alpha \int_0^\infty u^{\alpha – 1}\tfrac{e^{-uz}}{1 – e^{ – u}}du \end{gathered}$$

First we establish the hint, from

##\psi _n ( z )= {{\left( { – 1} \right)}^{n + 1}}n!\int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – \prod_{q = 1}^{n + 1} \lambda _q \right)^{ – 1}\prod_{k = 1}^{n + 1} \lambda _k^{z – 1}d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n+1##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _{n+1}}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_{n+1}}} \right)}}} \right| = \tfrac{d{y_{n+1}} \ldots d{y_1}}{{{y_{n}} \cdots {y_1}}}$$

$$\begin{gathered}\psi _{ \alpha } ( z )= \left( – 1 \right) ^{\alpha + 1} \Gamma ( \alpha +1 )\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }}\int_a^x \int_a^{y_1} \cdots \int_a^{y_n} \left( 1 – y_{n+1}\right) ^{ – 1}y_{n+1}^{z}\tfrac{d{y_{n+1}} \ldots d{y_1}}{y_{n+1} \cdots y_1} \\ = \left( – 1 \right) ^{\alpha + 1}\Gamma ( \alpha +1 )\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x^z}{1 – x} \right) \\ = \left( – 1 \right) ^{\alpha + 1}\tfrac{\Gamma ( \alpha +1 )}{ \Gamma ( \alpha ) } \mathop {\lim }\limits_{x \to {1^ – }}\int_0^\infty u^{\alpha – 1}\tfrac{x^ze^{ -z u}}{1 – x e^{ – u}}du \\ = \left( – 1 \right) ^{\alpha + 1} \alpha \int_0^\infty u^{\alpha – 1}\tfrac{e^{ -z u}}{1 – e^{ – u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}\psi _{\alpha}\left( z \right) = (-1)^{\alpha +1} \alpha \int_0^\infty u^{\alpha – 1}\tfrac{e^{-uz}}{1 – e^{ – u}}du \\ = (-1)^{\alpha +1} \alpha\sum_{k=0}^\infty \int_0^\infty u^{\alpha – 1}e^{-(z+k)}du \\ = (-1)^{\alpha +1} \sum_{k=0}^\infty\tfrac{\alpha}{(z+k)^\alpha} \int_0^\infty \omega ^{\alpha – 1}e^{-\omega}d\omega \\ = (-1)^{\alpha +1} \sum_{k=0}^\infty\tfrac{\Gamma (\alpha +1)}{(z+k)^\alpha} \\ \end{gathered}$$

c) Polylogarithm of Order n: ##{\text{L}}{{\text{i}}_n}\left( z \right): = \sum\limits_{q = 1}^\infty  {\frac{{{z^q}}}{{{q^n}}} = z\int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – z\prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}{\text{L}}{{\text{i}}_\alpha}\left( z \right) = z\mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x}{1-zx}} \right) =\tfrac{z}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 – ze^{ – u}}du \end{gathered}$$

First we establish the hint, from

##{\text{L}}{{\text{i}}_n}\left( z \right) = z\int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – z\prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1} \prod\limits_{k = 1}^n {d{\lambda _k}}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}{\text{L}}{{\text{i}}_{ \alpha }}\left( z \right)= z{\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n}{ 1 – zy_{n}} \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = z{\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1 – xz} \right) \\ = \tfrac{z}{\Gamma ( \alpha )}{ \mathop {\lim }\limits_{x \to {1^ – }}} \int_0^\infty u^{\alpha – 1}\tfrac{xe^{ – u}}{1 – xz e^{ – u}}du \\ = \tfrac{z}{\Gamma ( \alpha )} \int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – z e^{ – u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}{\text{L}}{\text{i}_\alpha}\left( z \right) = \tfrac{z}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 – ze^{ – u}}du \\ = \tfrac{z}{\Gamma (\alpha )}\sum_{k=0}^\infty z^k\int_0^\infty u^{\alpha – 1} e^{-(k+1)u}du \\ = \tfrac{z}{\Gamma (\alpha )}\sum_{k=0}^\infty \tfrac{z^k}{(k+1)^\alpha }\int_0^\infty \omega ^{\alpha – 1} e^{-\omega}d\omega \\ = \sum_{k=1}^\infty \tfrac{z^{k}}{k^\alpha}\\ \end{gathered}$$

d) Hurwitz Zeta Function: ##\zeta \left( {n,z} \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {q + z} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {\lambda _k^{z – 1}d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\zeta \left( {n,z} \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( {\tfrac{x^z}{1-x}} \right) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{-zu}}{1 – e^{ – u}}du \end{gathered}$$

First we establish the hint, from

##\zeta \left( n,z \right)  = \int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – \prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^n \lambda _k^{z – 1}d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}\zeta \left( \alpha , z \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n^{z}}{ 1 – y_{n}} \cdot \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = {\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x^z}{1 – x } \right) \\ = \tfrac{1}{\Gamma ( \alpha )} { \mathop {\lim }\limits_{x \to {1^ – }}}\int_0^\infty u^{\alpha – 1}\tfrac{x^{z}e^{ -zu}}{1 – x e^{ – u}}du \\ = \tfrac{1}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – zu}}{1 – e^{ -u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}\zeta \left( \alpha ,z \right) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{-zu}}{1 – e^{ – u}}du \\ =\tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty \int_0^\infty u^{\alpha – 1}e^{-(z+k)u}du \\ =\tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty \tfrac{1}{(z+k)^\alpha}\int_0^\infty \omega ^{\alpha – 1}e^{-\omega}d\omega \\ =\sum_{k=0}^\infty \tfrac{1}{(z+k)^\alpha} \\ \end{gathered}$$

e) Dirichlet Beta Function: ##\beta \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{{{{\left( { – 1} \right)}^q}}}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\beta\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1+x^2} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -2 u}}du \end{gathered}$$

First we establish the hint, from

##\beta \left( n \right) = \int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 + \prod\limits_{q = 1}^n \lambda _q^2  \right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}\beta \left( \alpha \right) = {\mathop {\lim }\limits_{a \to {0^ + }} }{\mathop {\lim }\limits_{x \to {1^ – }}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n}{ 1 + y_{n}^2} \cdot \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = {\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1 + x^2 } \right) \\ = \tfrac{1}{\Gamma ( \alpha )} { \mathop {\lim }\limits_{x \to {1^ – }}}\int_0^\infty u^{\alpha – 1}\tfrac{xe^{ -u}}{1 + x^2 e^{ – 2 u}}du \\ = \tfrac{1}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 + e^{ -2u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}\beta\left( \alpha \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -2 u}}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty \int_0^\infty u^{\alpha – 1} e^{-(2k+1)u}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty \tfrac{1}{(2k+1)^\alpha}\int_0^\infty \omega ^{\alpha – 1} e^{-\omega}d\omega \\ \end{gathered}$$

f) Dirichlet Eta Function: ##\eta \left( n \right): = \sum\limits_{q = 1}^\infty  {\frac{{{{\left( { – 1} \right)}^{q – 1}}}}{{{q^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 + \prod\limits_{q = 1}^n {{\lambda _q}} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\eta\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1+x} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -u}}du \end{gathered}$$

First we establish the hint, from

##\eta \left( n \right) = \int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 + \prod\limits_{q = 1}^n \lambda _q \right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}\beta \left( \alpha \right) = {\mathop {\lim }\limits_{a \to {0^ + }} }{\mathop {\lim }\limits_{x \to {1^ – }}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n}{ 1 + y_{n}} \cdot \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = {\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1 + x} \right) \\ = \tfrac{1}{\Gamma ( \alpha )} { \mathop {\lim }\limits_{x \to {1^ – }}}\int_0^\infty u^{\alpha – 1}\tfrac{xe^{ -u}}{1 + xe^{ -u}}du \\ = \tfrac{1}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 + e^{ -u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}\eta\left( \alpha \right) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 +e^{ -u}}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty (-1)^k\int_0^\infty u^{\alpha – 1} e^{-(k+1)u}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty \tfrac{(-1)^k}{(k+1)^\alpha }\int_0^\infty u^{\alpha – 1} e^{-(k+1)u}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=1}^\infty \tfrac{(-1)^{k-1}}{k^\alpha }\int_0^\infty \omega ^{\alpha – 1} e^{-\omega}d\omega \\ = \sum_{k=1}^\infty \tfrac{(-1)^{k-1}}{k^\alpha } \\ \end{gathered}$$

g) Dirichlet Lambda Function: ##\lambda \left( n \right): = \sum\limits_{q = 0}^\infty  {\frac{1}{{{{\left( {2q + 1} \right)}^n}}} = \int_0^1 {\int_0^1 { \cdots \int_0^1 {{{\left( {1 – \prod\limits_{q = 1}^n {\lambda _q^2} } \right)}^{ – 1}}\prod\limits_{k = 1}^n {d{\lambda _k}} } } } }##

Hint: $$\begin{gathered}\lambda\left( \alpha \right) = \mathop {\lim }\limits_{a \to {0^ + }} \mathop {\lim }\limits_{x \to {1^ – }} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1-x^2} \right) =\tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 -e^{ -2 u}}du \end{gathered}$$

First we establish the hint, from

##\lambda \left( n \right): = \int_0^1 \int_0^1  \cdots \int_0^1 \left( 1 – \prod\limits_{q = 1}^n \lambda _q^2 \right) ^{ – 1}\prod\limits_{k = 1}^n d{\lambda _k}##

Apply the Hypercube to Simplex transformation of Example 5.4, namely ##{y_k} = \prod_{i = 1}^k {\lambda _i} ,k = 1,2, \ldots ,n##,

$$\left| {\tfrac{{\partial \left( {{\lambda _1}, \ldots ,{\lambda _n}} \right)}}{{\partial \left( {{y_1}, \ldots ,{y_n}} \right)}}} \right| = \tfrac{{d{y_n} \ldots d{y_1}}}{{{y_{n – 1}} \cdots {y_1}}}$$

$$\begin{gathered}\beta \left( \alpha \right) = {\mathop {\lim }\limits_{a \to {0^ + }} }{\mathop {\lim }\limits_{x \to {1^ – }}}\int_a^x \int_a^{y_1} \cdots \int_a^{y_{n-1}} \tfrac{y_n}{ 1 – y_{n}^2} \cdot \tfrac{d{y_n} \ldots d{y_1}}{{y_n}{y_{n – 1}} \cdots {y_1}} \\ = {\mathop {\lim }\limits_{a \to {0^ + }}}{ \mathop {\lim }\limits_{x \to {1^ – }}} \,_a^{eH}I_x^\alpha \left( \tfrac{x}{1 – x^2} \right) \\ = \tfrac{1}{\Gamma ( \alpha )} { \mathop {\lim }\limits_{x \to {1^ – }}}\int_0^\infty u^{\alpha – 1}\tfrac{xe^{ -u}}{1 – xe^{-2u}}du \\ = \tfrac{1}{\Gamma ( \alpha )}\int_0^\infty u^{\alpha – 1}\tfrac{e^{ – u}}{1 – e^{ -2u}}du \\ \end{gathered}$$

And that establishes the hint.

$$\begin{gathered}\lambda\left( \alpha \right) = \tfrac{1}{\Gamma (\alpha )}\int_0^\infty u^{\alpha – 1} \tfrac{e^{-u}}{1 -e^{ -2 u}}du \\ = \tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty\int_0^\infty u^{\alpha – 1} e^{-(2k+1)u}du \\ =\tfrac{1}{\Gamma (\alpha )}\sum_{k=0}^\infty\tfrac{1}{(2k+1)^\alpha}\int_0^\infty \omega ^{\alpha – 1} e^{-\omega}\omega \\ = \sum_{k=0}^\infty\tfrac{1}{(2k+1)^\alpha} \\ \end{gathered}$$

## Footnotes

Footnotes
[1] I used as a guide for bivariate induction proofs: https://www.mathblog.dk/proof-method-multidimensional-induction/

[2] This theorem can be found in almost any undergraduate text on real analysis, for example, Principles of Mathematical Analysis by Rudin, 3rd edition, p. 321

## Works Cited

Works Cited

As for the reference material I have used the text Special Functions by Askey, Andrews, and Roy which covers much of the theorems here outlined.

## Appendix: Louiville’s Generalization of Dirichlet Integrals – Proofs

Though it is not used in this work, this generalization of Dirichlet’s result is proved as the stronger theorem of Louisville, namely

#### Theorem 2.0A: Louiville’s Generalization of Dirichlet Integrals

If ##t,{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_t^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq t} } \right\}##, and if F is continuous on [0,t], then

$$\begin{gathered}\iint {\mathop \cdots \limits_{V_t^n} \int {F\left[ {\sum\limits_{i = 1}^n {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}} \\ \cdot \int_{u = 0}^t {{u^{\sum\limits_{p = 1}^n {\left( {\frac{{{\gamma _p}}}{{{\beta _p}}}} \right)} – 1}}F ( u )du} \end{gathered}$$

##### Proof:

The proof is by induction. Let ##{J_n}## denote the left-hand side of the above equality.

(i) ##V_t^1: = \left\{ {{z_1} \in \mathbb{R}|0 \leq {{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}} \leq t} \right\}, {J_1} = \int_{{z_1} = 0}^{{\alpha _1}{t^{\frac{1}{{{\beta _1}}}}}} {F\left[ {{{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}}} \right]z_1^{{\gamma _1} – 1}d{z_1}} ##; let ##{z_1} = {\alpha _1}{u^{\frac{1}{{{\beta _1}}}}} \Rightarrow d{z_1} = \frac{{{\alpha _1}}}{{{\beta _1}}}{u^{\frac{1}{{{\beta _1}}}\, – 1}}du## so that

##{J_1} = \int_{u = 0}^t {{{\left( {{\alpha _1}{u^{\tfrac{1}{{{\beta _1}}}}}} \right)}^{{\gamma _1} – 1}}F\left( u \right)\left( {\frac{{{\alpha _1}}}{{{\beta _1}}}{u^{\tfrac{1}{{{\beta _1}}}\, – 1}}} \right)du}  = \frac{{\alpha _1^{{\gamma _1}}}}{{{\beta _1}}}\int_{u = 0}^t {{u^{\tfrac{{{\gamma _1}}}{{{\beta _1}}}}}^{ – 1}F\left( u \right)du} ##.

(ii) Assume the theorem holds for some fixed n. Then we have

$$\begin{gathered}V_t^{n + 1}: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^{n + 1} {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq t} } \right\} \\ = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}|0 \leq {z_1} \leq {\alpha _1}{t^{\tfrac{1}{{{\beta _1}}}}},0 \leq {z_j} \leq {\alpha _j}{{\left[ {t – \sum\limits_{k = 1}^{j – 1} {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}}} } \right]}^{\tfrac{1}{{{\beta _j}}}}}\forall j \geq 2} \right\} \\ \end{gathered}$$

$$\begin{gathered}{J_{n + 1}} = \iint {\mathop \cdots \limits_{V_t^{n + 1}} \int {F\left[ {\sum\limits_{i = 1}^{n + 1} {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]\prod\limits_{\lambda = 1}^{n + 1} {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_{n + 1}} \ldots d{z_2}d{z_1}} } \\ = \int_{{z_1} = 0}^{{\alpha _1}{t^{\tfrac{1}{{{\beta _1}}}}}} {\int_{{z_2} = 0}^{{\alpha _2}{{\left[ {t – {{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}}} \right]}^{\tfrac{1}{{{\beta _2}}}}}} { \cdots \int_{{z_{n + 1}} = 0}^{{\alpha _{n + 1}}{{\left[ {t – \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}}} } \right]}^{\tfrac{1}{{{\beta _{n + 1}}}}}}} {F\left[ {\sum\limits_{i = 1}^{n + 1} {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]\prod\limits_{\lambda = 1}^{n + 1} {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_{n + 1}} \ldots d{z_2}d{z_1}} } } \\ \end{gathered}$$

Apply the (n+1) transformation equations ##{z_m} = {\alpha _m}y_m^{\tfrac{1}{{{\beta _m}}}}\forall m \Rightarrow d{z_{n + 1}} \ldots d{z_2}d{z_1} = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{{\alpha _m}}}{{{\beta _m}}}y_m^{\tfrac{1}{{{\beta _m}}}\, – 1}} \right)} d{y_{n + 1}} \ldots d{y_2}d{y_1}## , to get

$$\begin{gathered}{J_{n + 1}} = \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{{y_{n + 1}} = 0}^{t – \sum\limits_{k = 1}^n {{y_k}} } {F\left( {\sum\limits_{i = 1}^{n + 1} {{y_i}} } \right)\prod\limits_{\lambda = 1}^{n + 1} {\left[ {{{\left( {{\alpha _\lambda }y_\lambda ^{\tfrac{1}{{{\beta _\lambda }}}}} \right)}^{{\gamma _\lambda } – 1}}\frac{{{\alpha _\lambda }}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{1}{{{\beta _\lambda }}}\, – 1}} \right]} d{y_{n + 1}} \ldots d{y_2}d{y_1}} } } \\ = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)} \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{{y_{n + 1}} = 0}^{t – \sum\limits_{k = 1}^n {{y_k}} } {F\left( {\sum\limits_{i = 1}^{n + 1} {{y_i}} } \right)\prod\limits_{\lambda = 1}^{n + 1} {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} d{y_{n + 1}} \ldots d{y_2}d{y_1}} } } \\ \end{gathered}$$

Substitute ##{y_{n + 1}} = {y_n}\left( {1 – u} \right){u^{ – 1}} \Rightarrow d{y_{n + 1}} =  – {y_n}{u^{ – 2}}du## to arrive at

$${J_{n + 1}} = \prod_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)} \\ \cdot \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{{y_n} = 0}^{t – \sum_{k = 1}^{n – 1} {{y_k}} } {\int_{u = {y_n}{{\left( {1 – \sum_{i = 1}^{n – 1} {{y_i}} } \right)}^{ – 1}}}^{1} {F\left( {\frac{{{y_n}}}{u} + \sum\limits_{i = 1}^n {{y_i}} } \right){{\left( {1 – u} \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}{u^{ – \,\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}y_{n + 1}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}\prod_{\lambda = 1}^n {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} dud{y_\lambda} } } } }$$

Switch the order of integration from ##dud{y_n} \ldots d{y_2}d{y_1}## to ##d{y_n}dud{y_{n – 1}} \ldots d{y_2}d{y_1}## and we have

$${J_{n + 1}} = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)} \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{u = 0}^1 {\int_{{y_n} = 0}^{u\left( {1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } \right)} {F\left( {\frac{{{y_n}}}{u} + \sum\limits_{i = 1}^n {{y_i}} } \right){{\left( {1 – u} \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}{u^{ – \,\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}y_{n + 1}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}\prod_{\lambda = 1}^n {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} d{y_n}du \ldots d{y_2}d{y_1}} } } }$$

Substitute ##{y_n} = uY \Rightarrow d{y_n} = Ydu## to obtain

\begin{gathered}{J_{n + 1}} = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)} \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{u = 0}^1 {\int_{Y = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {F\left( {Y + \sum\limits_{i = 1}^n {{y_i}} } \right){{\left( {1 – u} \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}{u^{\tfrac{{{\gamma _n}}}{{{\beta _n}}}\, – 1}}{Y^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}\cdot\prod\limits_{\lambda  = 1}^{n – 1} {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} dYdu \ldots d{y_2}d{y_1}} } } }  \\ = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)\int_{u = 0}^1 {{{\left( {1 – u} \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}{u^{\tfrac{{{\gamma _n}}}{{{\beta _n}}}\, – 1}}du} } \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{Y = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {F\left( {Y + \sum\limits_{i = 1}^n {{y_i}} } \right){Y^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}\prod\limits_{\lambda  = 1}^{n – 1} {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} dY \ldots d{y_2}d{y_1}} } }  \\ = \prod\limits_{m = 1}^{n + 1} {\left( {\frac{{\alpha _m^{{\gamma _m}}}}{{{\beta _m}}}} \right)\frac{{\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)}}{{\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}} \int_{{y_1} = 0}^t {\int_{{y_2} = 0}^{t – {y_1}} { \cdots \int_{Y = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {F\left( {Y + \sum\limits_{i = 1}^n {{y_i}} } \right){Y^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}\, – 1}}\prod\limits_{\lambda  = 1}^{n – 1} {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} dY \ldots d{y_2}d{y_1}} } }  \\ \end{gathered}

Apply the n transformation equations:

$${y_m} = {\left( {\frac{{{z_m}}}{{{\alpha _m}}}} \right)^{{\beta _m}}}\forall m \leq \left( {n – 1} \right) \Rightarrow d{y_{n – 1}} \ldots d{y_2}d{y_1} = \prod\limits_{m = 1}^\infty {\left[ {\frac{{{\alpha _m}}}{{{\beta _m}}}{{\left( {\frac{{{z_m}}}{{{\alpha _m}}}} \right)}^{{\beta _m} – 1}}} \right]} d{z_{n – 1}} \ldots d{z_2}d{z_1}$$

$$Y = {\left( {\frac{{{z_n}}}{{{\alpha _n}}}} \right)^{{\beta _n}}} \Rightarrow dY = \frac{{{\beta _n}}}{{{\alpha _n}}}{\left( {\frac{{{z_n}}}{{{\alpha _n}}}} \right)^{{\beta _n} – 1}}d{z_n}$$

and the integral becomes

\begin{gathered}{J_{n + 1}} = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)}}{{\alpha _n^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}{\beta _{n + 1}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{z_1} = 0}^{{\alpha _1}{t^{\tfrac{1}{{{\beta _1}}}}}} {\int_{{z_2} = 0}^{{\alpha _2}{{\left[ {t – {{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}}} \right]}^{\tfrac{1}{{{\beta _2}}}}}} { \cdots \int_{{z_n} = 0}^{{\alpha _n}{{\left[ {t – \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}}} } \right]}^{\tfrac{1}{{{\beta _n}}}}}} {F\left[ {\sum\limits_{i = 1}^n {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]z_n^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}\prod\limits_{\lambda  = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } }  \\ = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)}}{{\alpha _n^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}{\beta _{n + 1}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int {\int {\mathop  \cdots \limits_{V_t^n} \int {F\left[ {\sum\limits_{i = 1}^n {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]z_n^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}\prod\limits_{\lambda  = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } }  \\ \end{gathered}

which by hypothesis

\begin{gathered}{J_{n + 1}} = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)}}{{\alpha _n^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}{\beta _{n + 1}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}} \cdot {{\prod\limits_{q = 1}^{n – 1} {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} \left( {\frac{{\alpha _n^{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}} \right)} \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^{n – 1} {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} \left( {\frac{{\alpha _n^{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}} \right)} {\Gamma \left( {1 + \sum\limits_{k = 1}^{n – 1} {\left( {\frac{{{\gamma _k}}}{{{\beta _k}}}} \right) + \tfrac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^{n – 1} {\left( {\frac{{{\gamma _k}}}{{{\beta _k}}}} \right) + \tfrac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}} } \right)}} \\ \cdot \int_{u = 0}^t {{u^{\sum\limits_{p = 1}^n {\left( {\frac{{{\gamma _p}}}{{{\beta _p}}}} \right)}  + \tfrac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}\, – 1}}F\left( u \right)du}  \\ = {{\prod\limits_{q = 1}^{n + 1} {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/ {\vphantom {{\prod\limits_{q = 1}^{n + 1} {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^{n + 1} {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^{n + 1} {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}} \cdot \int_{u = 0}^t {{u^{\sum\limits_{p = 1}^{n + 1} {\left( {\frac{{{\gamma _p}}}{{{\beta _p}}}} \right)}  – 1}}F\left( u \right)du}  \\ \end{gathered}

the required result.

#### Corollary 2.0B: Louiville’s Generalization of Dirichlet Integrals (Modified Domain)

If ##{t_1},{t_2},{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_{{t_1},{t_2}}^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,{t_1} \leq \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq {t_2}} } \right\}##, and if F is continuous on ##\left[ {{t_1},{t_2}} \right]## then

$$\begin{gathered}\iint {\mathop \cdots \limits_{V_{{t_1},{t_2}}^n} \int {F\left[ {\sum\limits_{i = 1}^n {{{\left( {\frac{{{z_i}}}{{{\alpha _i}}}} \right)}^{{\beta _i}}}} } \right]\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}} \\ \cdot \int_{u = {t_1}}^{{t_2}} {{u^{\sum\limits_{p = 1}^n {\left( {\frac{{{\gamma _p}}}{{{\beta _p}}}} \right)} – 1}}F(u)du} \\ \end{gathered}$$

A result due to Dirichlet is given by

#### Theorem 2.1: Dirichlet Integrals

If ##{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and $${V^n}: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq 1} } \right\}$$

then

##\iint {\mathop  \cdots \limits_{{V^n}} \int {\prod\limits_{\lambda  = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {{\prod\limits_{q = 1}^n {\left[{\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/ {\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}##, where ##\Re \left[ w \right]## denotes the real part of w.

##### Proof:

The proof is here obtained as a trivial corollary of Theorem 2.0A.

#### Corollary 2.2: Dirichlet Integrals (Modified Domain 1)

If ##t,{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_t^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq t} } \right\}##, then

$$\iint {\mathop \cdots \limits_{V_t^n} \int {\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {t^{\sum\limits_{p = 1}^n {\frac{{{\gamma _p}}}{{{\beta _p}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}$$

#### Corollary 2.3: Dirichlet Integrals (Modified Domain 2)

If ##{t_1},{t_2},{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_{{t_1},{t_2}}^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,{t_1} \leq \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq {t_2}} } \right\}##, then

$$\iint {\mathop \cdots \limits_{V_{{t_1},{t_2}}^n} \int {\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = \left( {t_2^{\sum\limits_{p = 1}^n {\frac{{{\gamma _p}}}{{{\beta _p}}}} } – t_1^{\sum\limits_{r = 1}^n {\frac{{{\gamma _r}}}{{{\beta _r}}}} }} \right){{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}$$

#### Corollary 2.4: Content Integral

If ##{\alpha _p},{\beta _q} > 0\forall p,q## and ##{V^n}: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq 1} } \right\}##, then

##{\text{Content}}\left( {{V^n}} \right): = \iint {\mathop  \cdots \limits_{{V^n}} \int {d{z_n} \ldots d{z_2}d{z_1}} } = {{\prod\limits_{q = 1}^n {\left[ {\frac{{{\alpha _p}}}{{{\beta _q}}}\Gamma \left( {\frac{1}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{{\alpha _p}}}{{{\beta _q}}}\Gamma \left( {\frac{1}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{1}{{{\beta _k}}}} } \right)}}##. Where content refers to hypervolume when ##n \geq 4##.

## Exercises 2 – Dirichlet Integrals

2.1) a) Show that ##{{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right.} {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}} = \frac{{{\beta _1}}}{{{\gamma _1}}}\prod\limits_{k = 1}^n {{\rm B}\left( {1 + \sum\limits_{j = 1}^k {\frac{{{\gamma _j}}}{{{\beta _j}}}} ,\frac{{{\gamma _{k + 1}}}}{{{\beta _{k + 1}}}}} \right)}##

1. b) Rewrite Theorem 2.1 using the result of part a).
2. c) Formulate an alternate proof of Theorem 2.1 by means of a product of beta functions.

9 replies
1. benorin says:
I have just this minute added complete solutions to Exercises in Section 4, I do now believe there’s only one problem whom does not have a solution provided now. Not bad!
2. benorin says:
The overhaul is through it’s main phase, though I’m still going to post some more Answers to Exercises – Section 5. If previously you glanced at this Insight Article and found it rather too lengthy for you taste, please give it another go being as I have streamlined this work with the recommendation that you simply scan through sections 1 & 2. Sections 3 & 4 however are important for the proofs, and section 5 is the main results in terms of fractional integral representations of the Lerch Transcendent family of special functions.

Looking for co-author of this note that it might be elevated to the status of a paper. If you are interested, please say so here in the comments or send me a private message.

3. benorin says:
Update: I added generous hints to Section 5 exercise 4) parts a) thru f), namely the desired fractional integral is stated: still remains to be shown that the fractional integral is equal to the series definition.
4. benorin says:
I just wanted to say that God gave me every bit of skills, inspiration, people who helped me flesh this note out over the years I worked on it. Some people that helped me after having worked on this note throughout college and finally having had my analysis prof Akemann from UCSB read what was the beginnings of this note, of which he said what I was working with wasn’t well defined, and some years past. A couple of years ago God told me the finish my paper (I learned the other day that this work is called a note, because papers are published-I had been calling it my paper for quite a long time tho) so I came to PF and I got help from @fresh_42 and @FactChecker with and actually quite a lot of other math people here on PF helped (that would be a long list but I will just say search threads started by me and containing the word paper and from these search results you can see all the help I got, also there’s an Insights and Blog dev sub forum which is hidden by default with a few more threads with several advisors who also helped me).

Sorry this turned into kind of long winded credit where credit is due. I do feel that God gets the credit here, having worked all things to this end.

I’m currently rewriting this note, and seek a co-author if any of you are interested? I will say that I’m neural diverse, and if you contact me being interested in being a co-author I will tell you point-blank what my issues are, I just don’t feel comfortable doing so here publicly as I’m not knowing how these types of issues are handled in math circles yet.

Sorry for the essay.

-Ben

5. benorin says:
A few days ago I added the rest of the solutions to the exercises in section 1 (at the end of the document).
6. benorin says:
I updated the Insight to include more exercises in sections 4 and 5 and added a few answers to section 1 at the end.
7. benorin says:
Yes, it is, I had used that particular definition of ##C_N^n## with the ##\leq n-1## (opposed to ##\leq n##) to accommodate and singular point of an a few integrands in section 4, whereas the latter would converge to ##\left[ 0,1\right]##. These two definitions cover all of the integrals in the text.
8. mathman says:
Do you have a specific question? ##\lim_{N\to \infty} C_N^n=[0,1)## is straightforward.