# The Analytic Continuation of the Lerch and the Zeta Functions

## Introduction

In this brief Insight article the analytic continuations of the Lerch Transcendent and Riemann Zeta Functions are achieved via the Euler’s Series Transformation (zeta) and a generalization thereof, the E-process (Lerch). Dirichlet Series is mentioned as a steppingstone. The continuations are given but not shown to be convergent by any means, though if you the reader would be interested in such write me in the comments and I may oblige with an update if I get around to it. Some basic complex analysis and (double) series manipulations are the only assumed knowledge herein.

## Euler’s Series Transformation and the E-Process

We wish to consider the supposed convergent alternating series ##\sum\limits_{k = 1}^\infty  {{{\left( { – 1} \right)}^{k – 1}}{a_k}} ## by use of the power series

$$f\left( x \right) = \sum\limits_{k = 1}^\infty {{{\left( { – 1} \right)}^{k – 1}}{a_k}} {x^k}\text{ (1.1) }$$

Which we require to converge for at least ## – 1 < x \leq 1## .  That we may effect the change of variable [cf. Knopp pg. 469]

$$x = \frac{y}{{1 – y}}\text{ (1.2) }$$

In (1.1), the following calculations are useful: recall that

$$\frac{{{d^k}}}{{d{y^k}}}{y^j} = \frac{{j!}}{{\left( {j – k} \right)!}}{y^{j – k}}\text{ (1.3)}$$

holds for integer ##k \leq j,j \in {\mathbb{Z}^ + }##,  also note that under the change of variables (1.2), we have

\begin{gathered} {x^k} = {\left( {\tfrac{y}{{1 – y}}} \right)^k} = {y^k}{\left( {\tfrac{1}{{1 – y}}} \right)^k} = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!}} \cdot \tfrac{{{d^{k – 1}}}}{{d{y^{k – 1}}}}\left( {\tfrac{1}{{1 – y}}} \right) = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!}} \cdot \tfrac{{{d^{k – 1}}}}{{d{y^{k – 1}}}}\left( {\sum\limits_{j = 0}^\infty  {{y^j}} } \right) \\ = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!}}\sum\limits_{j = k – 1}^\infty  {\tfrac{{j!}}{{\left( {j – k + 1} \right)!}}{y^{j – k + 1}}}  = \sum\limits_{j = k – 1}^\infty  {\left( {\begin{array}{*{20}{c}} j \\ {k – 1} \end{array}} \right){y^{j + 1}}}  \\ \end{gathered}

where it is required that ##\left| y \right| < 1## for convergence so that (1.1) becomes

\begin{gathered} f\left( {\tfrac{y}{{1 – y}}} \right) = \sum\limits_{k = 1}^\infty  {{{\left( { – 1} \right)}^{k – 1}}{a_k}{{\left( {\tfrac{y}{{1 – y}}} \right)}^k}}  = \sum\limits_{k = 1}^\infty  {{{\left( { – 1} \right)}^{k – 1}}{a_k}\sum\limits_{j = k – 1}^\infty  {\left( {\begin{array}{*{20}{c}}  j \\ {k – 1}  \end{array}} \right){y^{j + 1}}} }  \\  = \sum\limits_{k = 0}^\infty  {\sum\limits_{j = k}^\infty  {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}  j \\  k  \end{array}} \right){a_{k + 1}}{y^{j + 1}}} }  = \sum\limits_{j = 0}^\infty  {{y^{j + 1}}\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}  j \\  k  \end{array}} \right){a_{k + 1}}} }  \\ \end{gathered}

By (1.2), we have ##x = 1## when ##y = \tfrac{1}{2}##, and considering ##f\left( 1 \right) ## gives Euler’s Series Transformation, namely

$$\boxed{\sum\limits_{k = 1}^\infty {{{\left( { – 1} \right)}^{k – 1}}{a_k}} = \sum\limits_{j = 0}^\infty {\tfrac{1}{{{2^{j + 1}}}}\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right){a_{k + 1}}} } }\text{ (1.4)}$$

Note that Knopp makes judicious use of the forward difference operator notation in this context, not reproduced herein.

A generalization of Euler’s Series Transformation is the E-Process (short for Euler Process)  is derived similar to the Euler’s Series Transformation by considering slight alterations of (1.1) and (1.2). Consider now the function defined by formal power series

$$g\left( x \right) = \sum\limits_{k = 1}^\infty {{a_k}{x^k}}\text{ (1.5)}$$

which is supposed to converge for at least ## – 1 < x \leq 1##. For fixed ##q##, propose the change of variables

$$x = \frac{y}{{1 – qy}}\text{ (1.6)}$$

under which, and utilizing (1.3), we have

\begin{gathered} {x^k} = {\left( {\tfrac{y}{{1 – qy}}} \right)^k} = {y^k}\tfrac{1}{{{{\left( {1 – qy} \right)}^k}}} = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!{q^{k – 1}}}} \cdot \tfrac{{{d^{k – 1}}}}{{d{y^{k – 1}}}}\left( {\tfrac{1}{{1 – qy}}} \right) = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!{q^{k – 1}}}} \cdot \tfrac{{{d^{k – 1}}}}{{d{y^{k – 1}}}}\left( {\sum\limits_{j = 0}^\infty  {{q^j}{y^j}} } \right) \\ = \tfrac{{{y^k}}}{{\left( {k – 1} \right)!{q^{k – 1}}}}\sum\limits_{j = k – 1}^\infty  {\tfrac{{j!}}{{\left( {j – k + 1} \right)!}}{q^j}{y^{j – k + 1}}}  = \tfrac{1}{{{q^{k – 1}}}}\sum\limits_{j = k – 1}^\infty  {\left( {\begin{array}{*{20}{c}}  j \\  {k – 1} \end{array}} \right){q^j}{y^{j + 1}}}  \\ \end{gathered}

where for convergence it is required that ##\left| {qy} \right| < 1##; apply (1.6) to (1.5) and it becomes

\begin{gathered} g\left( {\tfrac{y}{{1 – qy}}} \right) = \sum\limits_{k = 1}^\infty  {{a_k}{{\left( {\tfrac{y}{{1 – qy}}} \right)}^k}}  = \sum\limits_{k = 1}^\infty  {\tfrac{{{a_k}}}{{{q^{k – 1}}}}\sum\limits_{j = k – 1}^\infty  {\left( {\begin{array}{*{20}{c}}  j \\ {k – 1} \end{array}} \right){q^j}{y^{j + 1}}} }  \\  = \sum\limits_{k = 1}^\infty  {\sum\limits_{j = k}^\infty  {\tfrac{{{a_{k + 1}}}}{{{q^k}}}\left( {\begin{array}{*{20}{c}}  j \\  k \end{array}} \right){q^j}{y^{j + 1}}} }  = \sum\limits_{j = 0}^\infty  {{q^j}{y^{j + 1}}\sum\limits_{k = 0}^j {\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right)\frac{{{a_{k + 1}}}}{{{q^k}}}} }  \text{     (1.7)}\\ \end{gathered}

By (1.6), we have ##x = 1## when ##y = \tfrac{1}{{1 + q}}##, upon which ##\left| {qy} \right| < 1 \Rightarrow \left| {\tfrac{q}{{1 + q}}} \right| < 1 \Rightarrow \Re \left[ q \right] > \tfrac{1}{2}##, so that equating ##f\left( 1 \right) ## as given (1.5) by (1.7) and with ##y = \tfrac{1}{{1 + q}}## yields the result (the E-Process)

$$\boxed{\sum\limits_{k = 1}^\infty {{a_k}} = \frac{1}{{1 + q}}\sum\limits_{j = 0}^\infty {{{\left( {\frac{q}{{1 + q}}} \right)}^j}\sum\limits_{k = 0}^j {\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right) \frac{{{a_{k + 1}}}}{{{q^k}}}} } }\text{ (1.8)}$$

Clearly, setting ##q = 1## in yields Euler’s Series Transformation for Non-Alternating Series.

## The Analytic Continuations of the Lerch Transcendent

The Lerch Transcendent is defined as

$$\Phi \left( {z,s,u} \right) = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{{{\left( {u + k} \right)}^s}}}}\text{ (2.1)}$$

for ##u \in \mathbb{C} – \left\{ {0, – 1, – 2, \ldots } \right\};s \in \mathbb{C}{\text{ when }}\left| z \right| < 1,{\text{ and }}\Re \left[ s \right] > 1{\text{ when }}\left| z \right| = 1##. The analytic continuation of (2.1) given by (Guillera and Sondow, 2005, pg 5) may be obtained from (1.8): put ##{a_k}: = \tfrac{{{z^k}}}{{{{\left( {k + u} \right)}^s}}}##, set ##q =  – z## and the restriction becomes ##\left| {\tfrac{{ – z}}{{1 – z}}} \right| < 1 \Rightarrow \Re \left[ z \right] < \tfrac{1}{2}##, this slightly modified E-process yields

$$\Phi \left( {z,s,u} \right) = \frac{1}{{1 – z}}\sum\limits_{j = 0}^\infty {{{\left( {\frac{{ – z}}{{1 – z}}} \right)}^j}\sum\limits_{k = 0}^j {\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right)\frac{{{{\left( { – 1} \right)}^k}}}{{{{\left( {k + u} \right)}^s}}}} } \text{ (2.2)}$$

the right-hand side of which converges for ##u \in \mathbb{C} – \left\{ {0, – 1, – 2, \ldots } \right\};s \in \mathbb{C},\Re \left[ z \right] < \tfrac{1}{2}##.

## Dirichlet Series and the Global Analytic Continuation of the Riemann Zeta Function

Let us define a Dirichlet series as a series of the form

$$\xi \left( s \right) = \sum\limits_{k = 1}^\infty {\frac{{{a_k}}}{{{k^s}}}} \text{ (3.1)}$$

where ##{a_k},s \in \mathbb{C}##; if ##{a_k}## is multiplicative, that is, if ##\forall n,m \in {\mathbb{Z}^ + },{a_{nm}} = {a_n}{a_m}##, then

$$\begin{gathered} \xi \left( s \right) + \sum\limits_{k = 1}^\infty {{{\left( { – 1} \right)}^k}\frac{{{a_k}}}{{{k^s}}}} = \sum\limits_{k = 1}^\infty {\frac{{{a_k}}}{{{k^s}}}} + \sum\limits_{k = 1}^\infty {{{\left( { – 1} \right)}^k}\frac{{{a_k}}}{{{k^s}}}} = 2\sum\limits_{k = 1}^\infty {\frac{{{a_{2k}}}}{{{{\left( {2k} \right)}^s}}}} \\ = {2^{1 – s}}\sum\limits_{k = 1}^\infty {\frac{{{a_2}{a_k}}}{{{k^s}}}} = {2^{1 – s}}{a_2}\xi \left( s \right) \\ \end{gathered}$$

let us denote the series thus obtained by analytic continuation by

$$\hat \xi \left( s \right) = {\left( {1 – {2^{1 – s}}{a_2}} \right)^{ – 1}}\sum\limits_{k = 1}^\infty {{{\left( { – 1} \right)}^k}\frac{{{a_k}}}{{{k^s}}}}\text{ (3.2)}$$

which is analogous to the typical analytic continuation of the Riemann zeta function to the half-plane ##\Re \left[ s \right] > 0##. Let us further suppose that this alternating series is convergent, now apply Euler’s Series Transformation (1.4) to ##\hat \xi \left( s \right) ##

$$\hat \xi \left( s \right) = {\left( {1 – {2^{1 – s}}{a_2}} \right)^{ – 1}}\sum\limits_{j = 0}^\infty {\tfrac{1}{{{2^{j + 1}}}}\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right)\frac{{{a_{k + 1}}}}{{{{\left( {k + 1} \right)}^s}}}} } \text{ (3.3)}$$

At once this gives the global analytic continuation of the Riemann zeta function, namely

$$\boxed{\forall s \in \mathbb{C} – \left\{ 1 \right\},\zeta \left( s \right) = {{\left( {1 – {2^{1 – s}}} \right)}^{ – 1}}\sum\limits_{j = 0}^\infty {\tfrac{1}{{{2^{j + 1}}}}\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right)\frac{1}{{{{\left( {k + 1} \right)}^s}}}} } } \text{ (3.4)}$$

for this put ##{a_k} = 1## (which is multiplicative). Immediate consequences of this are ##\zeta \left( 0 \right) =  – \tfrac{1}{2},\zeta \left( { – 1} \right) =  – \tfrac{1}{{12}}##, since

\begin{gathered} \zeta \left( 0 \right) =  – \sum\limits_{j = 0}^\infty  {\tfrac{1}{{{2^{j + 1}}}}\underbrace {\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}  j  \\  k  \end{array}} \right)} }_{ = 0{\text{ for }}j \geq 1}}  =  – \tfrac{1}{2} \hfill \\ \zeta \left( { – 1} \right) =  – \tfrac{1}{3}\sum\limits_{j = 0}^\infty  {\tfrac{1}{{{2^{j + 1}}}}\underbrace {\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}  j \\  k \end{array}} \right)\left( {k + 1} \right)} }_{ = 0{\text{ for }}n \geq 2}}  =  – \tfrac{1}{{12}} \hfill \\ \end{gathered}

where ##\forall N \in \mathbb{N},n > N \Rightarrow \sum\limits_{k = 0}^n {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}}  n \\  k  \end{array}} \right){{\left( {k + 1} \right)}^N}}  = 0## (which is (24) on pg. 6 of Guillera and Sondow, 2005) has been employed.

## Conclusion

Here they are,

$$\boxed{\forall s \in \mathbb{C} – \left\{ 1 \right\},\zeta \left( s \right) = {{\left( {1 – {2^{1 – s}}} \right)}^{ – 1}}\sum\limits_{j = 0}^\infty {\tfrac{1}{{{2^{j + 1}}}}\sum\limits_{k = 0}^j {{{\left( { – 1} \right)}^k}\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right)\frac{1}{{{{\left( {k + 1} \right)}^s}}}} } }$$

$$\boxed{ \forall u \in \mathbb{C} – \left\{ {0, – 1, – 2, \ldots } \right\},\forall s \in \mathbb{C},,\forall z \in \mathbb{C}{\text{ such that }}\Re \left[ z \right] < \tfrac{1}{2} \\ \Phi \left( z,s,u \right) =\frac{1}{1 – z}\sum\limits_{j = 0}^\infty {{{\left( {\frac{{ – z}}{{1 – z}}} \right)}^j}\sum\limits_{k = 0}^j {\left( {\begin{array}{*{20}{c}} j \\ k \end{array}} \right) \frac{{{{\left( { – 1} \right)}^k}}}{{{{\left( {k + u} \right)}^s}}}} } }$$

## Citations

Guillera, J. and Sondow, J., 2005. DOUBLE INTEGRALS AND INFINITE PRODUCTS FOR SOME CLASSICAL CONSTANTS VIA ANALYTIC CONTINUATIONS OF LERCH’S TRANSCENDENT. [online] Arxiv.org. Available at: < https://arxiv.org/pdf/math/0506319.pdf  > [Accessed 23 April 2020].