TITLE: Integrating $\int^\infty_ 0\frac{x^2}{e^{x^2} - 1}\,\mathrm dx$
QUESTION [5 upvotes]: How can I integrate
$$\int^{\infty}_ 0 \frac{x^2}{{e^{x^2}} - 1}\,\mathrm{d}x?$$
I have tried using the Gaussian Integral Formula and Integration by parts, but have made no progress so far.

REPLY [2 votes]: As was mentioned in the comments, I'll have fun generalizing the result to
$$I(s)=\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx \qquad s\in\mathbb{C}, \text{Re }(s)> 1$$
We'll expand the denominator into a geometric series before using the Gamma function to evaluate the final integral:
\begin{align}
I(s)&=\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx \\ 
&=\int_0^{+\infty} \frac{x^s e^{-x^2}}{1-e^{-x^2}}\,dx \\
&=\int_0^{+\infty} x^s e^{-x^2} \sum_{n=0}^{+\infty} e^{-nx^2},dx \\
&=\sum_{n=0}^{+\infty}\int_0^{+\infty} x^s e^{-x^2} e^{-nx^2},dx \\
&=\sum_{n=0}^{+\infty}\int_0^{+\infty} x^s e^{-x^2(n+1)},dx \\
&=\sum_{n=1}^{+\infty}\int_0^{+\infty} x^s e^{-nx^2}dx \qquad u=x^2\\ 
&=\frac{1}{2}\sum_{n=1}^{+\infty}\int_0^{+\infty} u^{s/2} e^{-nu} \frac{du}{\sqrt{u}} \\
&=\frac{1}{2}\sum_{n=1}^{+\infty}\int_0^{+\infty} u^{(s-1)/2} e^{-nu}\,du \\
&=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right)\sum_{n=1}^{+\infty} \frac{1}{n^{(s+1)/2}}\\ \\
I(s)&=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right) \zeta\left(\frac{s+1}{2}\right)
\end{align}
Thus, for $s\in\mathbb{C}, \text{Re }(s)>1$, 
\begin{align}
\boxed{\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right) \zeta\left(\frac{s+1}{2}\right)}
\end{align}