Integral representation for polylogarithm with a negative index

Question

From the power series definition of the polylogarithm and from the integral representation of the Gamma function it is easy to show that: \begin{equation} Li_{s}(z) := \sum\limits_{k=1}^\infty k^{-s} z^k = \frac{z}{\Gamma(s)} \int\limits_0^\infty \frac{\theta^{s-1}}{e^\theta-z} d \theta \end{equation} The identity holds whenever $Re(s) > 0$. Now my question is twofold.

Firstly, how do we analytically continue that function to the area $Re(s) <0$? Clearly this must be possible because it was already Riemann who found a corresponding reflection formula by deforming the integration contour to the complex plane and evaluating that integral both in a clock-wise and in a anti-clockwise direction.

My second question would be how do we compute two dimensional functions of that kind. To be precise I am interested in quantities like this:

\begin{equation} Li_{s_1,s_2}^{(\xi_1,\xi_2)}(z_1,z_2) := \sum\limits_{1 \le k_1 < k_2 < \infty }(k_1+\xi_1)^{-s_1} (k_2+\xi_2)^{-s_2} z_1^{k_1} z_2^{k_2-k_1} \end{equation} Clearly if both $Re(s_1) >0$ and $Re(s_2) >0$ the quantity above has a following integral representation: \begin{equation} Li_{s_1,s_2}^{(\xi_1,\xi_2)}(z_1,z_2) = \frac{z_1 z_2}{\Gamma(s_1) \Gamma(s_2)} \int\limits_{{\mathbb R}_+^2} \frac{\theta_1^{s_1-1} \theta_2^{s_2-1} e^{-\theta_1 \xi_1-\theta_2 \xi_2}}{\left(e^{\theta_1+\theta_2}-z_1\right)\left(e^{\theta_2}-z_2\right)} d\theta_1\theta_2 \end{equation} However how do I compute the quantity if any of the real parts of the $s$-parameters becomes negative?

Answer 1

Let us answer the first question namely how do we analytically continue the polylogarithm to negative values of $s$. In here we just use the method devised by Riemann in his 1859 paper. He applied that method to the Zeta function and we will generalize it to the poly-logarithm. The integral representation of the poly-logarithm reads: \begin{equation} Li_s(z) = \frac{z}{\Gamma(s)} \cdot \frac{\imath}{2 \sin(\pi s)} \cdot \int_{\infty}^\infty \frac{(-\theta)^{s-1}}{e^\theta-z} d \theta \end{equation} where the integration contour runs in positive direction from plus infinity surrounds the origin, but no other singularity of the integrand, and then returns back to plus infinity. Since the contour integral itself, i.e. without the prefactors equals just $\left( (-\imath)^{s-1} + \imath^{s-1}\right)$ times the original integral along the positive half-axis, the definition above reduces to the old definition when $Re(s) > 0 $.

Now we can also evaluate the integral in negative direction and then it equals the sum of residues about all poles $\left\{ \log(z) + 2 \pi \imath n \right\}_{n=-\infty}^\infty$. Since the integral about such pole reads $(-\log(z)-2 \pi \imath n)^{s-1} \cdot 1/z \cdot (-2 \pi \imath)$ we easily establish the following relation: \begin{eqnarray} &&Li_s(z) = \Gamma(1-s) \cdot \\ && \left[(-\log(z))^{s-1}+(2 \pi)^{s-1} \left(\imath^{s-1} \zeta(1-s,1+\imath \frac{\log(z)}{2 \pi}) + (-\imath)^{s-1} \zeta(1-s,1-\imath \frac{\log(z)}{2 \pi})\right)\right] \\ && = \Gamma(1-s) \cdot \\ && \left[(-\log(z))^{s-1}+(2 \pi)^{s-1} \sum\limits_{n=0}^\infty \binom{s-1}{n} \zeta(n+1-s) \cdot 2 \cos(\frac{\pi}{2} (s-1+n)] \cdot \left(\frac{\log(z)}{2 \pi}\right)^n \right] \end{eqnarray} where $\zeta$ is the Hurwitz zeta function.

Answer 2

I. question anwser

Let's use (Infinite series as integral representation), which is working on divergent series too.

$\displaystyle \sum_{k=x}^{\infty} f (k)=-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt $

Let's $f (k)=k^{-s}z^k \rightarrow F (k)=-ln^{s-1} \left (1/z \right)\Gamma (1-s,ln (1/z)k)$

Now we can just substitute

$\displaystyle \sum_{k=1}^{\infty} k^{-s}z^k=2\pi ln^{s-1} \left (1/z\right)\int_{-\infty}^{\infty} \frac { \Gamma \left (1-s,(\frac {1}{2}+it)ln (1/z) \right)}{(e^{\pi t}+e^{-\pi t})^2}dt $

Eventually you can use generalised Abel-Plana formula with assumtion that series at infinity is convergent (Convergence of definite integral and its analytic continuation)

$ \displaystyle \sum_{k=x}^{\infty}f (k)=-F(x)+\frac {1}{2}f (x)+\int_{0}^{\infty} \frac{f (x+it)-f (x-it)}{e^{2\pi t}-1} dt. $

So you get some nicer formula.

$\displaystyle \sum_{k=1}^{\infty} k^{-s}z^k= \Gamma(1-s, -\ln(z))\ln^{s-1} (1/z)+\frac {z}{2}+i\int_{0}^{\infty} \frac{ (1+it)^{-s}z^{1+it}- (1-it)^{-s}z^{1-it} }{e^{2\pi t}-1} dt. $

I could't add image on mobile. Here is link to photo that proves my secound equation is true. $Li_s (z)=z \Phi (z,s,1) $ (https://wikimedia.org/api/rest_v1/media/math/render/svg/7ae19d36a078bb68987acd5304b8a82c2e09c158)