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.

Thanks in advance!

  • 1
    $\begingroup$ Looks like an application of Ramanujan's Master Theorem, coupled with the reflection formula for the Gamma function. $\endgroup$ – nospoon Sep 29 '18 at 14:52
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ 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 :) $\endgroup$ – 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


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

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