# Show that $\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}$

I have come across the following integral while going over this list (Problem $$35$$)

$$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}~~~~s>0, \alpha\in(0,1)$$ where $$\operatorname{Li}_s(x)$$ denotes the Polylogarithm Function.

Using the series expansion of $$\operatorname{Li}_s(-x)$$ yields

\begin{align} \int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx&=\int_0^{\infty}\frac1{x^{\alpha+1}}\left[\sum_{n=1}^{\infty}\frac{(-x)^n}{n^s}\right]\mathrm dx\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}\int_0^{\infty}\frac{x^n}{x^{\alpha+1}}\mathrm dx\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}\int_0^{\infty}x^{n-\alpha-1}\mathrm dx \end{align}

One can easily see the problems concerning the convergence of the last integral. Also, I am not even sure whether it is possible to change the order of summation and integration in this case or not.

Another approach is based on an integral representation of $$\operatorname{Li}_s(-x)$$ so that the given integral becomes

\begin{align} \int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx&=\int_0^{\infty}\frac1{x^{\alpha+1}}\left[\frac1{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{e^t/(-x)-1}\mathrm dt\right]\mathrm dx\\ &=-\frac1{\Gamma(s)}\int_0^{\infty}\int_0^{\infty}\frac{t^{s-1}}{x^{\alpha}(e^t+x)}\mathrm dx\mathrm dt\\ \end{align}

From hereon, I do not know how to proceed. Since the solution reminds me of Euler's Reflection Formula it might be possible to reshape the integral in terms of the Gamma Function somehow.

I am asking for a whole evaluation of the given integral. I did not found anything closely related to this question but correct me if I am wrong.

• Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function. – nospoon Sep 29 '18 at 14:52
• @nospoon Is it possible to use this Theorem here? I also thought about a potential connection to the Mellin Transform but hence the $x$-term has the form $x^{-\alpha-1}$ instead of $x^{\alpha-1}$ I do not know whether this possible or not. – mrtaurho Sep 29 '18 at 14:58

You're just one step away: for $$b>0$$, we have $$\int_0^\infty {\frac{1}{{{x^\alpha }(b + x)}}dx} = {b^{ - \alpha }}\int_0^\infty {\frac{{{x^{ - \alpha }}}}{{1 + x}}dx} = \frac{{{b^{ - \alpha }}\pi }}{{\sin \alpha \pi }}$$ so $$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}dx=-\frac{1}{\Gamma(s)}\int_0^{\infty}\int_0^{\infty}\frac{t^{s-1}}{x^{\alpha}(e^t+x)}dxdt = - \frac{\pi }{{\Gamma (s)\sin \alpha \pi }}\int_0^\infty {{t^{s - 1}}{e^{ - \alpha t}}dt}$$

The series expansion of $$\text{Li}_s(-x)$$ only converges for $$|x| < 1$$, so your first method is illegitimate.

• First of all thank you for your answer. I guessed for myself that I went wrong somewhere concerning the convergence of the series expansion. Furhermore I was not aware of this integral identity at the moment. I think it can be shown using the Beta Function hence I remember some kind of identity similiar to this? But nevertheless I am glad you provided an asnwer :) – mrtaurho Sep 29 '18 at 15:17

I have finally figured out how to use Ramanujans Master Theorem in this case.

Ramanujans Master Theorem

Let $$f(x)$$ be an analytic function with a MacLaurin Expansion of the form $$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}x^{s-1}f(x)\mathrm dx=\Gamma(s)\phi(-s)$$

Lets get back to the given integral. The series representation of $$\operatorname{Li}_s(-x)$$ is given by $$\displaystyle\sum_{k=1}^{\infty}\frac{(-x)^k}{k^s}$$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $$\displaystyle\phi(k)=\frac{\Gamma(k+1)}{k^s}$$. Now applying the Mellin Transform with $$s=-\alpha$$ yields to

$$\int_0^{\infty}x^{-\alpha-1}\operatorname{Li}_s(-x)\mathrm dx=\Gamma(-\alpha)\phi(\alpha)=\Gamma(-\alpha)\frac{\Gamma(\alpha+1)}{\alpha^s}$$

This term can be simplified by using Eulers Reflection Formula with $$z=\alpha+1$$ which finally leads to

$$\frac1{\alpha^s}\Gamma(\alpha+1)\Gamma(-\alpha)=\frac1{\alpha^s}\frac{\pi}{\sin(\pi(\alpha+1))}=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi\alpha)}$$

• RMT always finds a way... +1 :) – clathratus Mar 20 '19 at 2:36