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

Recently 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.

I tried to use the series expansion of $$\operatorname{Li}_s(-x)$$ which yields to

\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}

But one can easily see the problems concerning the convergence of the last integral. Furthermore I am not sure whether it is possible to interchange the order of summation and integration in this case or not.

Another approach would be to use the an integral represantation 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}

But from hereon I have no clue how to proceed. Since the solution reminds me of Euler's Reflection Formula it is maybe possible somehow to reshaphe the integral in terms of the Gamma Function.

Therefore I am asking for a whole evaluation of the given integral. I did not found anything closely connected to this question but you can correct me if I have overseen something.

• 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 at 2:36