Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need help with solving this integral: $$\int_0^\infty x^{-\frac{3}{2}}\ \text{Li}_{\sqrt{2}}(-x)\ dx,$$ where $\text{Li}_{s}(z)$ is the polylogarithm.

share|cite|improve this question
up vote 12 down vote accepted

$$\int_0^\infty x^{-\frac{3}{2}}\ \text{Li}_{\sqrt{2}}(-x)\ \mathrm dx=-2^{\sqrt{2}}\pi.$$


Use formula (3) from here to get an integral representation of the polylogarithm: $$\text{Li}_s(-x)=-\frac{1}{\Gamma(s)}\int_0^\infty\frac{k^{s-1}}{\frac{e^k}{x}-1}\mathrm dk.$$ Then, changing the order of integration, $$\int_0^\infty x^{-p}\ \text{Li}_s(-x)\mathrm dx=-\frac{1}{\Gamma(s)}\int_0^\infty\int_0^\infty\frac{x^{-p}\,\ k^{s-1}}{\frac{e^k}{x}-1}\mathrm dx\ \mathrm dk=\\\frac{\pi}{\Gamma(s)\sin \pi p}\int_0^\infty e^{k(1-p)}\ k^{s-1}\mathrm dk=\frac{\pi}{(p-1)^s \sin \pi p}$$

share|cite|improve this answer
(+1) A nice solution! – Sangchul Lee May 22 '13 at 19:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.