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I want to show that $$\operatorname{Li}_2(z)=z\int_{0}^{\infty}\frac{x}{e^x-z}dx$$

It is the integral of the Bose–Einstein distribution in dilogarithm case.

Thank you!

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1 Answer 1

We have \begin{align*} z\int_0^{+\infty}\frac x{e^x-z}dx &=z\int_0^{+\infty}\frac{xe^{-x}}{1-ze^{-x}}dx\\ &=z\int_0^{+\infty}xe^{-x}\sum_{n=0}^{+\infty}z^ne^{-nx}dx\\ &=z\sum_{n=0}^{+\infty}z^n\int_0^{+\infty}xe^{-(n+1)x}dx\\ \end{align*} (to invert the series and the integral we need the dominated convergence theorem, which works here since $|z|<1$). Since $$\int_0^{+\infty}xe^{-(n+1)x}dx=-\frac 1{n+1}\left[xe^{-(n+1)x}\right]_0^{+\infty}+\frac 1{n+1}\int_0^{+\infty}e^{-(n+1)x}dx=\frac 1{(n+1)^2},$$ we get $$z\int_0^{+\infty}\frac x{e^x-z}dx=z\sum_{n=0}^{+\infty}\frac{z^n}{(n+1)^2}=\sum_{j=1}^{+\infty}\frac{z^j}{j^2}=Li_2(z).$$

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