Integral Representations of Some Special Functions

A Path to Fractional Integral Representations of Some Special Functions

0. Introduction

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. Another reference text, I cite Theory and Applications of Infinite Series by Knopp. 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 the integrals

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

might be original material as I have never seen it my reading though there is nothing new under the Sun.

Summary

Section 1

Gives a glimpse of the gamma and beta functions. As for special functions, the gamma function is the least special 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, others as well.

Section 2

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

Is dedicated mostly to defining a sequence of sets that point-wise converge 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 formulae for certain special functions represented as integrals over the unit hypercube evaluated in a unique fashion.

Section 5

Expands the previous section’s special functions represented as multiple integrals to fractional integrals which provide analytic continuations of the prior sections identities to complex orders of integration, in particular the Hadamard fractional integral operator is employed to this end.

1. A Treatment of the Gamma and Beta Functions

“Read Euler: he is our master in everything!” –PS Laplace

The Gamma Function

Definition 1.0: The Factorial

##z!: = \prod\limits_{q = 1}^z q ##

It was Leonhard Euler who, at the age of 22, was the first to extend the domain of z! (the factorial) from non-negative integer arguments to all complex arguments less the negative integers. The analytic continuation of z! was likely suggested to Euler by either his then friend, colleague, and roommate Daniel Bernoulli, or by another colleague of Euler’s at the St. Petersburg Academy of Science: Christian Goldbach. Insofar as one can be certain, the later is to whom, Euler, in his letter of October 13th, 1729, disclosed his solution.

One way to achieve Euler’s result is to observe that ##\forall z,\lambda  \in {\mathbb{Z}^*}: = {\mathbb{Z}^ + } \cup \left\{ 0 \right\}## one may write

\begin{eqnarray*} z! &=& \prod\limits_{q = 1}^z q  = \left( {\prod\limits_{q = 1}^z q } \right)\underbrace {\mathop {\lim }\limits_{\lambda  \to \infty } \left( {\prod\limits_{j = z + 1}^{z + \lambda } j } \right){{\left( {\prod\limits_{\rho  = z + 1}^{z + \lambda } \rho  } \right)}^{ – 1}}}_{ = 1}\\ &=& \mathop {\lim }\limits_{\lambda  \to \infty } \left( {\prod\limits_{j = 1}^{z + \lambda } j } \right){\left[ {\prod\limits_{k = 1}^\lambda  {\left( {z + k} \right)} } \right]^{ – 1}} = \mathop {\lim }\limits_{\lambda  \to \infty } \left( {\prod\limits_{k = 1}^\lambda  k } \right)\left( {\prod\limits_{j = \lambda  + 1}^{z + \lambda } j } \right)\prod\limits_{k = 1}^\lambda  {\left( {\frac{1}{{z + k}}} \right)}\\&=& \mathop {\lim \,}\limits_{\lambda  \to \infty } \lambda !\prod\limits_{q = 1}^z {\left( {\lambda  + q} \right)} \prod\limits_{k = 1}^\lambda  {\left( {\frac{1}{{z + k}}} \right)}  = \mathop {\lim \,}\limits_{\lambda  \to \infty } \lambda !\left[ {{\lambda ^z}\prod\limits_{q = 1}^z {\left( {\frac{{\lambda  + q}}{\lambda }} \right)} } \right]\prod\limits_{k = 1}^\lambda  {\left( {\frac{1}{{z + k}}} \right)} \\ &=& \mathop {\lim \,}\limits_{\lambda  \to \infty } \lambda !{\lambda ^z}\prod\limits_{k = 1}^\lambda  {\left( {\frac{1}{{z + k}}} \right)} : = \Gamma \left( {z + 1} \right) \end{eqnarray*}

 yielding the so-called Euler limit form of the gamma function, namely

Definition 1.1: Euler Limit Form of the Gamma Function

(Euler 1729): ## \forall z \notin {\mathbb{Z}^ – } \cup \left\{ 0 \right\},\;\Gamma \left( z \right): = \mathop {\lim }\limits_{\lambda  \to \infty } \frac{{\lambda !{\lambda ^{z – 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda  – 1} \right)}}## , a meromorphic function of the complex variable z which possesses simple poles at non-positive integer arguments.

From Definition 1.1 it follows that a difference equation satisfied by the gamma function is

Equation 1.2: Difference Equation for the Gamma Function

##\forall z \notin {\mathbb{Z}^ – }\cup\left\{ 0\right\},\;\Gamma \left( {z+1}\right) = z\Gamma \left( z \right)##

Another elementary property of the gamma function whose proof is also left as an exercise for the reader is given by

Equation 1.3: Relation Between Gamma Function and the Factorial

##\forall n \in {\mathbb{Z}^*},\,\Gamma (n + 1) = n!## .

That the Euler product form of the gamma function is often taken to be the definition thereof notwithstanding[1], in this work the author shall have the occasion to call it

Theorem 1.1: Euler Product Form of the Gamma Function

(Euler 1729, and Gauss 1811): ##\forall z \notin {\mathbb{Z}^ – } \cup \left\{ 0 \right\},## ##\Gamma \left( z \right) = \frac{1}{z}\prod\limits_{\rho  = 1}^\infty  {\left[ {{{\left( {1 + \frac{1}{\rho }} \right)}^z}{{\left( {1 + \frac{z}{\rho }} \right)}^{ – 1}}} \right]} ##.

Proof:

\begin{eqnarray*}\Gamma (z) &:=& \mathop {\lim }\limits_{\lambda\to\infty }\frac{\lambda !\lambda ^{z – 1}}{z (z+1)\cdots\left( z + \lambda  – 1 \right)} = \mathop {\lim }\limits_{\lambda\to\infty }\frac{\lambda !{\lambda ^{z – 1}}}{z\left( z+1\right)\cdots\left( z \lambda  – 1\right)}\cdot\underbrace {\mathop {\lim }\limits_{\lambda\to\infty}\frac{\lambda}{z +\lambda }}_{ = 1}\\ &=& \mathop {\lim}\limits_{\lambda \to\infty}\frac{\lambda !\lambda ^z}{z (z+1)\cdots (z+\lambda )} =\frac{1}{z}\mathop {\lim }\limits_{\lambda \to \infty } \;{\lambda ^z}\prod\limits_{\rho  = 1}^\lambda  \left( \frac{\rho }{z + \rho } \right) \\ &=& \frac{1}{z}\mathop {\lim }\limits_{\lambda  \to \infty}\;\left[\frac{\lambda (\lambda +1) !}{ (\lambda +1)\lambda !}\right]^{z}\prod\limits_{\rho  = 1}^{\lambda}\left(\frac{\rho }{z + \rho } \right) = \frac{1}{z}\mathop {\lim }\limits_{\lambda  \to \infty } \;\left[ \frac{\lambda }{\lambda  + 1}\prod\limits_{k = 1}^{\lambda} \left(\frac{k + 1}{k}\right)\right]^{z}\prod\limits_{\rho  = 1}^\lambda \left( \frac{\rho }{z+ \rho } \right)\\ &=& \frac{1}{z}\mathop {\lim }\limits_{\lambda  \to \infty } \;{\left( {1 – \frac{1}{{\lambda  + 1}}} \right)^z}\prod\limits_{\rho  = 1}^\lambda  {\left[ {{{\left( {1 + \frac{1}{\rho }} \right)}^z}\left( {\frac{\rho }{{z + \rho }}} \right)} \right]}  = \frac{1}{z}\prod_{\rho = 1}^\infty  {\left[ {{{\left( {1 + \frac{1}{\rho }} \right)}^z}{{\left( {1 + \frac{z}{\rho }} \right)}^{ – 1}}} \right]}\end{eqnarray*}

Not but thirteen weeks after the first, Christian Goldbach received another letter from Euler, this one dated January 8th, 1730, containing yet another definition of the gamma function, being convergent for arguments with non-negative real parts, was originally known as an Eulerian Integral of the Second Kind, was later dubbed the Gamma Integral by Gauss, yet here is humbly retitled

Theorem 1.2: Gamma Function Integral

(Euler 1730):  ## \Gamma ( z ) = \int_{0}^{\infty} e^{ – t}t^{z – 1} dt , \, \Re \left[ z \right] > 0 ##

Proof:

\begin{eqnarray*} \Gamma \left( z \right) &:=& \mathop {\lim }\limits_{\lambda  \to \infty } \frac{{\lambda !{\lambda ^{z – 1}}}}{{z\left( {z + 1} \right) \cdots \left( {z + \lambda  – 1} \right)}} \\&=& \mathop {\lim }\limits_{\lambda  \to \infty } \frac{\lambda ! \lambda ^{z – 1}}{z ( z + 1 ) \cdots ( z + \lambda  – 2)}\int_{0}^{1} (1-x)^{z + \lambda – 2} dx\end{eqnarray*}

Now, integrate by parts to get

\begin{eqnarray*}\Gamma ( z) &=& \mathop {\lim }\limits_{\lambda  \to \infty } \frac{\lambda ! \lambda ^{z – 1}}{z (z + 1) \cdots ( z + \lambda  – 2 )}\left\{ \left[ \left. x ( 1 – x) ^{z + \lambda  – 2} \right| \right._{x = 0}^1 + \left( {z + \lambda  – 2} \right)\int_{0}^{1} x (1 – x)^{z + \lambda  – 3}dx\right\} \\ &=& \mathop {\lim }\limits_{\lambda  \to \infty } \frac{\lambda !\lambda ^{z – 1}}{z ( z + 1) \cdots ( z + \lambda  – 3 ) } \int_{0}^{1} x (1-x) ^{z + \lambda  – 3} dx\end{eqnarray*}

In general, k iterations of integration by parts gives

$$\Gamma (z) = \mathop {\lim }\limits_{\lambda  \to \infty } \frac{\lambda !\lambda ^{z – 1}}{z(z + 1) \cdots ( z + \lambda  – k – 2 ) k!}\int_{0}^1 ( 1 – x)^{z + \lambda  – k – 2} x^k dx $$

in particular, ##\left( {\lambda  – 1} \right)##  iterations of integration by parts gives

$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda  \to \infty } \, \, \lambda ^{z – 1} \int_{0}^{1} ( 1 – x)^{z – 1} x^{\lambda  – 1} dx $$

Substitute ##{x^\lambda } = y \Rightarrow \lambda {x^{\lambda  – 1}}dx = dy## , to get

$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda  \to \infty }  \lambda ^{z – 1} \int_{0}^{1} \left( 1 – y^{\frac{1}{\lambda }} \right) ^{z – 1} dy $$

Set ##\lambda  = \frac{1}{\eta }## , so that ##\eta  \to {0^ + }##  as ##\lambda  \to \infty ## and

$$\Gamma (z) = \mathop {\lim }\limits_{\eta  \to {0^{+}}} \int_{0}^{1} \left( \frac{1 – y^{\eta }}{\eta } \right) ^{z – 1} dy \,\mathop  = \limits^H \,\;\int_{0}^{1} {\log }^{z – 1}\left( \frac{1}{y} \right) dy $$

where ##\mathop  = \limits^H ##  denotes the use of L’Hospital’s Rule. Substitute ##y={e^{-t}} \Rightarrow dy = – {e^{-t}}dt## , to get

$$\Gamma \left( z \right) = \;\int_{0}^\infty  e^{ – t}t^{z – 1} dt$$

and the theorem is demonstrated.

Weierstrass took as the definition of the gamma function it’s canonical infinite product representation, the so-called Weierstrass product form of the gamma function. The desired representation of the gamma function is here obtained as a corollary to the Weierstrass Factor Theorem‡the proof of which shall not be reproduced here[2].

Theorem 1.3: Weierstrass Factor Theorem

(Weierstrass): Let ##f(z)## be an entire (i.e. everywhere analytic) function with simple zeroes at ## z = a_1,a_2,a_3,\ldots##  where ##0 < \left| {{a_1}} \right| < \left| a_2\right| < \left| a_3\right| <  \ldots##  and ##\mathop {\lim }\limits_{M \to \infty}\;{a_M} =\infty##, then

$$f\left( z \right) = f\left( 0 \right){e^{\frac{{f’\left( 0 \right)}}{{f\left( 0 \right)}}z}}\prod\limits_{k = 1}^\infty  {\left[ {\left( {1 – \frac{z}{{{a_k}}}} \right){e^{\frac{z}{{{a_k}}}}}}\right]}$$

The Weierstrass product form of the gamma function is then given by

Corollary 1.4: Weierstrass Product Form of the Gamma Function

(Weierstrass): ##\frac{1}{{\Gamma \left( z \right)}} = z{e^{\gamma z}}\prod\limits_{\lambda  = 1}^\infty  {\left[ {\left( {1 + \frac{z}{\lambda }} \right){e^{ – \frac{z}{\lambda }}}} \right]} ## where ##\gamma ## is Euler’s Constant.[3]

Proof:

Let ##f\left( z \right) = \frac{1}{{\Gamma \left( {z + 1} \right)}}##  so that f(z) is an entire function with simple zeroes at ##z = {a_k}: =  – k, \forall k \in {\mathbb{Z}^ + }##  satisfying the hypotheses necessary to invoke Theorem 1.2, which yields

$$\frac{1}{{\Gamma \left( {z + 1} \right)}} = {e^{f’\left( 0 \right)z}}\prod\limits_{k = 1}^\infty  {\left[ {\left( {1 + \frac{z}{k}} \right){e^{ – \frac{z}{k}}}} \right]} $$

Set ##z=1## in the above formula and take natural logarithms of the result to determine

\begin{eqnarray*} f^{\prime} (0) &=& \sum_{k = 1}^\infty  \left[ \frac{1}{k} – \log \left( 1 + \frac{1}{k} \right) \right] \\&=& \mathop {\lim }\limits_{M \to \infty } \,\sum_{k = 1}^M \left[ \frac{1}{k} + \log ( k )  – \log ( k + 1) \right] \\&=& \mathop {\lim }\limits_{M \to \infty } \,\left[ H_M  – \log ( M + 1) \right] +\log 1\\&=& \mathop {\lim }\limits_{M \to \infty } \,\left[ H_M – \log (M + 1)  \right] + \underbrace{\mathop {\lim }\limits_{M \to \infty } \log \left( 1 + \frac{1}{M} \right) }_{ = \log 1}\\&=& \mathop {\lim }\limits_{M \to \infty } \,\left( H_M- \log M \right) = :\gamma \end{eqnarray*}

Where ##H_M## is the ##\text{M}^{th}## harmonic number and  theorem is proved upon applying Equation 1.1 and replacing ##f’\left( 0 \right)##  with ##\gamma ## , which is Euler’s Constant. Euler’s constant to four decimal places is

$$\gamma : = \mathop {\lim }\limits_{M \to \infty } \left[ {\sum\limits_{k = 1}^M {\left( {\frac{1}{k}} \right) – \ln M} } \right] = 0.5772 \ldots $$

 

The Beta Function

Definition 1.5: The Beta Function

The Beta function is ## B (x,y) : = \frac{\Gamma (x) \Gamma ( y )}{\Gamma ( x + y ) }##.

A product can be formulated for this function easily from theorem 1.1 (the reader may verify)

Theorem 1.6: The Infinite Product for the Beta Function

##B (x,y) = \frac{x + y}{xy}\prod_{\lambda  = 1}^\infty  \frac{1 + \frac{x + y}{\lambda }}{\left( 1 + \frac{x}{\lambda } \right)\left( 1 + \frac{y}{\lambda } \right) } ##.

Theorem 1.6:  A Beta Function Integral

##\forall x,y \in \mathbb{C}## such that ##\Re \left[ x \right] > 0 \wedge \Re \left[ y \right] > 0,## we have ##B ( x,y ) = \int_{0}^1 (1-t)^{x – 1} t^{y – 1} dt ##.

Proof:

The method of proof here is going to be fairly straightforward, we begin by taking the product of integrals of theorem 1.2 so we can then apply techniques of multiple integrals like changes of variables.

\begin{eqnarray*}\Gamma (x)\Gamma (y) &=& \left(\int_{ 0}^{\infty} e^{-t}t^{x-1}dt\right)\left(\int_{0}^{\infty}e^{-w}w^{y – 1}dw\right)\\ &=& \int_{0}^{\infty}\int_{0}^{\infty}e^{-(w+t)}w^{y-1}t^{x-1}dtdw\end{eqnarray*}

Now it’s a double integral, so we can apply the transformation equations ##t = {u^2},w = {v^2}## having Jacobian ##J\left( {u,v} \right) = {{t_u}} {{w_v}}-{{t_v}}{{w_u}}=( 2u)(2v )= 4uv## to get

$$\Gamma ( x )\Gamma ( y ) = 4\int_{0}^{\infty } \int_{0}^{\infty } e^{ – \left( u^2 + v^2 \right)} u^{2y – 1}v^{2x – 1}\, du dv  $$

seeing the ##x^2 +y^2## term is our cue to transform to polar to arrive at (this may be made rigorous by Squeeze Theorem)

\begin{eqnarray*}\Gamma ( x ) \Gamma ( y ) &=& 4\int_{0}^{\frac{\pi }{2}} \int_{0}^{\infty } e^{ – {r^2}}r^{2 ( x+y ) – 1} \cos ^{2y – 1} \theta \sin ^{2x – 1} \theta\, dr d\theta\\&=& 4\int_{0}^{\frac{\pi}{2}} \cos ^{ 2y – 1} \theta \sin ^{ 2x – 1} \theta d\theta  \int_{0}^{\infty } e^{ – {r^2} } r^{2 (x+y) -1} dr\end{eqnarray*}

and substitute ##z = r^2\Rightarrow\frac{dz}{2} = rdr## and the integral becomes

\begin{eqnarray*}\Gamma ( x ) \Gamma ( y ) &=& 2\int_{0}^{\frac{\pi}{2}} \cos ^{2y – 1}\theta \sin ^{2x – 1} \theta d\theta  \underbrace {\int_{0}^{\infty } e^{ – z}z^{ (x+y) – 1} dz}_{ = \Gamma (x+y) }\\&=& 2\Gamma (x+y)\int_{0}^{\frac{\pi }{2}}\cos ^{2y – 1}\theta\sin ^{2x – 1}\theta d\theta\end{eqnarray*}

now perform the change of variable ##s = {\cos ^2}\theta\Rightarrow ds = 2\cos\theta\sin\theta d\theta ## and we have

$$\Gamma (x)\Gamma (y)=\Gamma (x+y)\int_{0}^{1} (1-s)^{x-1}s^{y-1}ds$$

whence the theorem is proved.

Corollary 1.7: Another Beta Function Integral

##\forall x,y \in \mathbb{C} ## such that ## \Re \left[ x \right] > 0 \wedge \Re \left[ y \right] > 0, ##

$$B \left( {x,y} \right) = 2\int_{0}^{\frac{\pi}{2}} {\cos }^{2y – 1}\theta \, {\sin }^{2x – 1}\theta \, d\theta $$

Definition 1.8: The Polygamma Function

The polygamma function is denoted ##{\psi _n}## and is defined as

$${\psi _n}\left( z \right): = \frac{{{d^{n + 1}}}}{{d{z^{n + 1}}}}\log \Gamma \left( z \right) $$


Exercises 1 – Gamma and Beta Functions

1.1) Verify that Equation 1.1 follows from Definition 1.1.

1.2) a) Use the principal of mathematical induction and Definition 1.1 to establish Equation 1.2.

       b) Why do you think mathematicians chose to adopt the convention ##0!: = 1## ?

1.3) Use the Euler product form of the gamma function to show that ##\mathop {\lim }\limits_{N \to \infty } \frac{\Gamma ^n \left( 1 + \frac{k}{N} \right) }{ \Gamma \left( 1 + \frac{kn}{N} \right) } = 1##.

1.4) Show that ##\mathop {\rm{Res}} \limits_{z =  – k} \Gamma ( z ) = \frac{\left( -1\right) ^k}{k!} , \, \forall k \in \mathbb{Z}^{+} ##.

1.5) Obtain the formula ##\psi _n ( z ) = \sum\limits_{q = 0}^\infty  \frac{ (-1) ^{n + 1} n!}{ (z+q)^{n + 1}}## by taking as the definition of the gamma function either

         a) the Euler limit form, b) the Euler product form, c) the Weierstrass product form.

1.6) Evaluate ##\int_{\frac{1}{ 2}}^{1} \psi _1 ( z ) \, dz ##.

1.7) Choose your favorite product form of the gamma function and use it to prove ##\Gamma ( z ) \Gamma ( 1 – z ) = \frac{\pi }{\sin ( \pi x ) }##.

1.8) Prove that ##\Gamma ( z ) \zeta ( z ) = \int_{0}^{\infty}  \frac{u^{z – 1}}{e^u – 1}\, du ## for ##\Re \left[ z \right] > 1##, where ##\zeta ( z ) := \sum_{k = 1}^{\infty }  \frac{1}{k^z} ## is the Riemann-Zeta function.

1.9) Use partial fraction decomposition to show that ##\forall z \notin {\mathbb{Z}^ – } \cup \left\{ 0 \right\} ,##

$$\Gamma ( z ) = \mathop {\lim }\limits_{\lambda  \to \infty } \frac{\lambda ! \lambda ^{z – 1}}{z ( z + 1 ) \cdots ( z + \lambda  – 1 ) } = \mathop {\lim }\limits_{\lambda  \to \infty } \, \lambda ^{z – 1} \sum\limits_{k = 0}^{\lambda  – 1}  ( -1) ^k ( \lambda  – k ) \left. _{\lambda} C_{k} \right. ( z + k )^{ – 1}$$

where ##\left. _{\lambda} C_{k} \right. ## is a binomial coefficient.

 

2. The Integrals of Dirichlet; Louiville’s Generalization Thereof

 

Dirichlet Integrals

 

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 by induction. Let ##{I_n}## denote the integral on the left-hand side of the above equality.

 

(i) ##{V^1}: = \left\{ {{z_1} \in \mathbb{R}|0 \leq {z_1} \leq {\alpha _1}} \right\}##, ##{I_1} =\int_{{z_{1 = 0}}}^{{\alpha _1}} {z_1^{{\gamma _1} – 1}d{z_1}}  = \frac{{\alpha _1^{{\gamma _1}}}}{{{\gamma _1}}}\underbrace {\left[ {{{\frac{{{\gamma _1}}}{{{\beta _1}}}\Gamma \left( {\frac{{{\gamma _1}}}{{{\beta _1}}}} \right)} \mathord{\left/ {\vphantom {{\frac{{{\gamma _1}}}{{{\beta _1}}}\Gamma \left( {\frac{{{\gamma _1}}}{{{\beta _1}}}} \right)} {\Gamma \left( {1 +\frac{{{\gamma _1}}}{{{\beta _1}}}} \right)}}} \right. } {\Gamma \left( {1+ \frac{{{\gamma _1}}}{{{\beta _1}}}} \right)}}} \right]}_{ = 1} = {{\frac{{\alpha _1^{{\gamma _1}}}}{{{\beta _1}}}\Gamma \left( {\frac{{{\gamma _1}}}{{{\beta _1}}}} \right)} \mathord{\left/{\vphantom {{\frac{{\alpha _1^{{\gamma _1}}}}{{{\beta _1}}}\Gamma \left( {\frac{{{\gamma _1}}}{{{\beta _1}}}} \right)} {\Gamma \left( {1 + \frac{{{\gamma _1}}}{{{\beta _1}}}} \right)}}} \right. } {\Gamma \left( {1 + \frac{{{\gamma _1}}}{{{\beta _1}}}} \right)}}##

 

(ii) Assume the theorem holds for some fixed ##n \in {\mathbb{Z}^ + }## .

 

\begin{gathered}{V^{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 1} } \right\} \\ = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_{n + 1}}} \right) \in {\mathbb{R}^{n + 1}}|0 \leq {z_1} \leq {\alpha _1},0 \leq {z_j} \leq {\alpha _j}{{\left[ {1 – \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}

 

$${I_{n + 1}} = \iint {\mathop  \cdots \limits_{{V^{n + 1}}} \int {\prod\limits_{\lambda  = 1}^{n + 1} {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_{n + 1}} \ldots d{z_2}d{z_1}} } = \int_{{z_1} = 0}^{{\alpha _1}} {\int_{{z_2} = 0}^{{\alpha _2}{{\left[ {1 – {{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}}} \right]}^{\tfrac{1}{{{\beta _2}}}}}} { \cdots \int_{{z_{n + 1}} = 0}^{{\alpha _{n + 1}}{{\left[ {1 – \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}}} } \right]}^{\tfrac{1}{{{\beta _{n + 1}}}}}}} {\prod\limits_{\lambda  = 1}^{n + 1} {\left( {z_\lambda ^{{\gamma _\lambda } – 1}} \right)} d{z_{n + 1}} \ldots d{z_2}d{z_1}} } }$$

Apply the ##\left( {n + 1} \right)## 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}##

\begin{gathered}{I_{n + 1}} = \int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_{n + 1}} = 0}^{1 – \sum\limits_{k = 1}^n {{y_k}} } {\prod\limits_{\lambda  = 1}^{n + 1} {\left[ {{{\left( {{\alpha _\lambda }y_\lambda ^{\tfrac{1}{{{\beta _\lambda }}}}} \right)}^{{\gamma _\lambda } – 1}}\left( {\frac{{{\alpha _\lambda }}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{1}{{{\beta _\lambda }}} – 1}} \right)} \right]} d{y_{n + 1}} \ldots d{y_2}d{y_1}} } }  \\ = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}}}{{{\beta _{n + 1}}}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_{n + 1}} = 0}^{1 – \sum\limits_{k = 1}^n {{y_k}} } {y_{n + 1}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}} – 1}\prod\limits_{\lambda  = 1}^n {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_{n + 1}} \ldots d{y_2}d{y_1}} } }  \\ = \left( {\frac{{{\beta _{n + 1}}}}{{{\gamma _{n + 1}}}}} \right)\frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}}}{{{\beta _{n + 1}}}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_n} = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {\left[ {\left. {y_{n + 1}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}} \right|} \right._{{y_{n + 1}} = 0}^{1 – \sum\limits_{k = 1}^n {{y_k}} }\prod\limits_{\lambda  = 1}^n {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_n} \ldots d{y_2}d{y_1}} } }  \\ = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}}}{{{\gamma _{n + 1}}}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_n} = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {{{\left( {1 – \sum\limits_{k = 1}^n {{y_k}} } \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}\prod\limits_{\lambda  = 1}^n {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_n} \ldots d{y_2}d{y_1}} } }  \\ \end{gathered}

 

Noting that the variables ##{y_1},{y_2}, \ldots ,{y_{n – 1}}## are held constant when integrating with respect to ##{y_n}##, set

 

##{y_n} = \left( {1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } \right)x \Rightarrow d{y_n} = \left( {1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } \right)dx##

 

\begin{eqnarray*}{I_{n + 1}} &=& \frac{{\alpha _n^{{\gamma _n}}\alpha _{n + 1}^{{\gamma _{n + 1}}}}}{{{\beta _n}{\gamma _{n + 1}}}}\int_{x = 0}^1 {{x^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} – 1}}{{\left( {1 – x} \right)}^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}dx} \int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_{n – 1}} = 0}^{1 – \sum\limits_{r = 1}^{n – 2} {{y_r}} } {{{\left( {1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } \right)}^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}\prod\limits_{\lambda  = 1}^{n – 1} {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_{n – 1}} \ldots d{y_2}d{y_1}} } }  \\ &=& \frac{{\alpha _n^{{\gamma _n}}\alpha _{n + 1}^{{\gamma _{n + 1}}}}}{{{\beta _n}{\gamma _{n + 1}}}}\frac{{\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {1 + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}{{\Gamma \left( {1 + \tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_{n – 1}} = 0}^{1 – \sum\limits_{r = 1}^{n – 2} {{y_r}} } {{{\left( {1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} } \right)}^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}\prod\limits_{\lambda  = 1}^{n – 1} {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_{n – 1}} \ldots d{y_2}d{y_1}} } }  \\ &=& \frac{{\alpha _n^{{\gamma _n}}\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {1 + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}{{{\beta _n}{\gamma _{n + 1}}\Gamma \left( {1 + \tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_{n – 1}} = 0}^{1 – \sum\limits_{r = 1}^{n – 2} {{y_r}} } {\left[ {\left. {y_n^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}} \right|} \right._{{y_n} = 0}^{1 – \sum\limits_{k = 1}^{n – 1} {{y_k}} }\prod\limits_{\lambda  = 1}^{n – 1} {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_{n – 1}} \ldots d{y_2}d{y_1}} } }  \\ &=& \left( {\frac{{{\gamma _n}}}{{{\beta _n}}} + \frac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\frac{{\alpha _n^{{\gamma _n}}\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {1 + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}{{{\beta _n}{\gamma _{n + 1}}\Gamma \left( {1 + \tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_n} = 0}^{1 – \sum\limits_{r = 1}^{n – 1} {{y_r}} } {y_n^{\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}} – 1}\prod\limits_{\lambda  = 1}^{n – 1} {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_n} \ldots d{y_2}d{y_1}} } }  \\ &=& \left( {\frac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)\frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}{{{\gamma _{n + 1}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{y_1} = 0}^1 {\int_{{y_2} = 0}^{1 – {y_1}} { \cdots \int_{{y_n} = 0}^{1 – \sum\limits_{r = 1}^{n – 1} {{y_r}} } {y_n^{\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}\prod\limits_{\lambda  = 1}^n {\left( {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1}} \right)} d{y_n} \ldots d{y_2}d{y_1}} } }  \\ \end{eqnarray*}

 

Apply the ##n## transformation equations: ##{y_m} = {\left( {\frac{{{z_m}}}{{{\alpha _m}}}} \right)^{{\beta _m}}}\forall m \Rightarrow d{y_n} \ldots d{y_2}d{y_1} = \prod\limits_{m = 1}^n {\left[ {\frac{{{\beta _m}}}{{{\alpha _m}}}{{\left( {\frac{{{z_m}}}{{{\alpha _m}}}} \right)}^{{\beta _m} – 1}}} \right]} d{z_n} \ldots d{z_2}d{z_1}##

 

\begin{gathered}{I_{n + 1}} = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}{{{\beta _{n + 1}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}} + \tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \right)}}\int_{{z_1} = 0}^{{\alpha _1}} {\int_{{z_2} = 0}^{{\alpha _2}{{\left[ {1 – {{\left( {\frac{{{z_1}}}{{{\alpha _1}}}} \right)}^{{\beta _1}}}} \right]}^{\tfrac{1}{{{\beta _2}}}}}} { \cdots \int_{{z_n} = 0}^{{\alpha _{n + 1}}{{\left[ {1 – \sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}}} } \right]}^{\tfrac{1}{{{\beta _n}}}}}} {{{\left( {\frac{{{z_m}}}{{{\alpha _m}}}} \right)}^{\tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}\prod\limits_{\lambda  = 1}^n {\left[ {\frac{{\alpha _\lambda ^{{\gamma _\lambda }}}}{{{\beta _\lambda }}}{{\left( {\frac{{{z_\lambda }}}{{{\alpha _\lambda }}}} \right)}^{{\beta _\lambda }\left( {\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}} – 1} \right)}}\frac{{{\beta _\lambda }}}{{{\alpha _\lambda }}}{{\left( {\frac{{{z_\lambda }}}{{{\alpha _\lambda }}}} \right)}^{{\beta _\lambda } – 1}}} \right]} }}}\cdot d{z_n} \ldots d{z_2}d{z_1}  \\ = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \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)}}\iint {\mathop  \cdots \limits_{{V^n}} \int {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} {I_{n + 1}} = \frac{{\alpha _{n + 1}^{{\gamma _{n + 1}}}\Gamma \left( {\tfrac{{{\gamma _n}}}{{{\beta _n}}}} \right)\Gamma \left( {\tfrac{{{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}} \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 \tfrac{{\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( {\frac{{{\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)}  + \frac{{{\gamma _n} + \tfrac{{{\beta _n}{\gamma _{n + 1}}}}{{{\beta _{n + 1}}}}}}{{{\beta _n}}}} \right)}} \\  = {{\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)}} \\  \end{gathered}

the required result.

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##.

 

Louiville’s Generalization of Dirichlet Integrals

 

A generalization of Dirichlet’s result is given by the stronger theorem of Louiville, namely

 

Theorem 2.5: 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

 

##\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\left( u \right)du} ##.

 

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\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} = 0}^{t – \sum\limits_{k = 1}^{n – 1} {{y_k}} } {\int_{u = {y_n}{{\left( {1 – \sum\limits_{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\limits_{\lambda  = 1}^n {\left( {y_\lambda ^{\tfrac{{{\gamma _\lambda }}}{{{\beta _\lambda }}}\, – 1}} \right)} dud{y_n} \ldots d{y_2}d{y_1}} } } }##

 

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\limits_{\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 2.6: 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

 

##\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\left( u \right)du}##


Exercises II – 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.

2.2) Prove Corollary 2.2


3. 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_2^1 \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

##{\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)}}##


 4. 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 Reimann 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 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^{y – 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 IV:

 

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

4.2)

4.3)


5. 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 formulae 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

 

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

 

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 V – 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##.

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)##.

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:

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}} } } } }##

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}} } } } }##

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}} } } } }##

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^{y – 1}d{\lambda _k}} } } } }##

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}} } } } }##

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}} } } } }##

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}} } } } }##

 


Answers to Selected Exercises

 

Exercises 1

 

1.1) Verify that Equation 1.1 follows from Definition 1.1.

 

$$\Gamma \left( {z + 1} \right) = \mathop {\lim }\limits_{\lambda  \to \infty } \tfrac{{\lambda !{\lambda ^z}}}{{\left( {z + 1} \right)\left( {z + 2} \right) \ldots \left( {z + \lambda } \right)}} = \underbrace {\mathop {\lim }\limits_{\lambda  \to \infty } \tfrac{{\lambda z}}{{z + \lambda }}}_{ = z} \cdot \mathop {\lim }\limits_{\lambda  \to \infty } \tfrac{{\lambda !{\lambda ^{z – 1}}}}{{z\left( {z + 1} \right)\left( {z + 2} \right) \ldots \left( {z + \lambda  – 1} \right)}} = z\Gamma \left( z \right)$$

 

1.2) a) Use the principal of mathematical induction and Definition 1.1 to establish Equation 1.2.

The principle of mathematical induction affords that if one can demonstrate both (i) the truth of ##{P_1}## and (ii) for some fixed ##n \in {\mathbb{Z}^ + },{P_n} \Rightarrow {P_{n + 1}}##, then the truth of the infinite sequence of propositions ##{\left\{ {{P_k}} \right\}_{k \in {\mathbb{Z}^ + }}}## is established.

 

(i) $$\Gamma \left( 2 \right) = \Gamma \left( {1 + 1} \right) = 1 \cdot \Gamma \left( 1 \right) = \mathop {\lim }\limits_{\lambda  \to \infty } \tfrac{{\lambda !{\lambda ^0}}}{{1 \cdot \left( {1 + 1} \right) \ldots \left( {1 + \lambda  – 1} \right)}} = \mathop {\lim }\limits_{\lambda  \to \infty } \tfrac{{\lambda !}}{{\lambda !}} = 1$$

(ii) Assume that for some fixed ##n \in {\mathbb{Z}^ + },\Gamma \left( {n + 1} \right) = n!## is true, then ##\Gamma \left( {n + 2} \right) = \left( {n + 1} \right)\Gamma \left( {n + 1} \right) = \left( {n + 1} \right)n! = \left( {n + 1} \right)!##.

 

  1. b) Why do you think mathematicians chose to adopt the convention ##0!: = 1## ?

 

Substituting ##n = 0## into equation 1.2 yields ##0! = \Gamma \left( 1 \right) = 1##.

 

1.3) Use the Euler product form of the gamma function to show that ##\mathop {\lim }\limits_{N \to \infty } {{{\Gamma ^n}\left( {1 + \frac{k}{N}} \right)} \mathord{\left/ {\vphantom {{{\Gamma ^n}\left( {1 + \frac{k}{N}} \right)} {\Gamma \left( {1 + \frac{{kn}}{N}} \right)}}} \right. } {\Gamma \left( {1 + \frac{{kn}}{N}} \right)}} = 1## .

\begin{gathered}\mathop {\lim }\limits_{N \to \infty } \tfrac{{{\Gamma ^n}\left( {1 + \tfrac{k}{N}} \right)}}{{\Gamma \left( {1 + \tfrac{{kn}}{N}} \right)}} = \mathop {\lim }\limits_{N \to \infty } \tfrac{{{{\left( {1 + \tfrac{k}{N}} \right)}^{ – n}}\prod\limits_{\lambda  = 1}^\infty  {\tfrac{{{{\left( {1 + \tfrac{1}{\lambda }} \right)}^{n\left( {1 + \tfrac{k}{N}} \right)}}}}{{{{\left( {1 + \tfrac{{1 + \tfrac{{kn}}{N}}}{\lambda }} \right)}^n}}}} }}{{{{\left( {1 + \tfrac{{kn}}{N}} \right)}^{ – 1}}\prod\limits_{j = 1}^\infty  {\tfrac{{{{\left( {1 + \tfrac{1}{j}} \right)}^{1 + \tfrac{{kn}}{N}}}}}{{\left( {1 + \tfrac{{1 + \tfrac{{kn}}{N}}}{j}} \right)}}} }} \\ = \mathop {\lim }\limits_{N \to \infty } {\left( {1 + \tfrac{k}{N}} \right)^{ – n}}\left( {1 + \tfrac{{kn}}{N}} \right)\prod\limits_{\lambda  = 1}^\infty  {\left[ {\tfrac{{{{\left( {1 + \tfrac{1}{\lambda }} \right)}^{n – 1}}\left( {1 + \tfrac{{1 + \tfrac{{kn}}{N}}}{\lambda }} \right)}}{{{{\left( {1 + \tfrac{{1 + \tfrac{k}{N}}}{\lambda }} \right)}^n}}}} \right]}  \\ = \prod\limits_{\lambda  = 1}^\infty  {\left[ {{{\left( {1 + \tfrac{1}{\lambda }} \right)}^{n – 1}}\mathop {\lim }\limits_{N \to \infty } \tfrac{{\left( {1 + \tfrac{{1 + \tfrac{{kn}}{N}}}{\lambda }} \right)}}{{{{\left( {1 + \tfrac{{1 + \tfrac{k}{N}}}{\lambda }} \right)}^n}}}} \right]}  \\ = \prod\limits_{\lambda  = 1}^\infty  {\left[ {{{\left( {1 + \tfrac{1}{\lambda }} \right)}^{n – 1}}\tfrac{{\left( {1 + \tfrac{1}{\lambda }} \right)}}{{{{\left( {1 + \tfrac{1}{\lambda }} \right)}^n}}}} \right]}  = 1 \\ \end{gathered}

 

 

More Answers to be added later…


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 most any undergraduate text on real analysis, for example Principles of Mathematical Analysis by Rudin, 3rd edition, p. 321

[3] Thanks to FactChecker for this mapping from the thread: https://www.physicsforums.com/threads/i-need-a-mapping-from-the-unit-hypercube-0-1-n-to-a-given-simplex.985017/#post-6305550

Comment Thread

3 replies
  1. 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.

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