Integrating $\int^\infty_ 0\frac{x^2}{e^{x^2} - 1}\,\mathrm dx$

Question

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.

Answer 1

Using that

$$\frac1{e^{x^2}-1}=\sum_{n=1}^\infty e^{-nx^2}$$

it follows that

$$\int_0^\infty \frac{x^2}{e^{x^2}-1}\,dx =\sum_{n=1}^\infty\underbrace{\int_0^\infty e^{-nx^2}x^2\,dx}_{=\sqrt{\pi}/(4n^{3/2})} =\frac{\sqrt\pi}4\sum_{n\geq1}\frac1{n^{3/2}} =\frac{\sqrt\pi}4\zeta\left(\frac32\right)$$

Edit: the calculation of

$$\int_0^\infty e^{-nx^2}x^2\,dx$$

can be done using the classical result $\int_0^\infty e^{-n x^2}\,dx=\sqrt{\pi}/(2\sqrt{n})$ with $n>0$. Indeed,

$$\int_0^\infty e^{-nx^2}x^2\,dx =-\int_0^\infty \frac{d}{dn}e^{-nx^2}\,dx =-\frac{d}{dn}\frac{\sqrt{\pi}}{2\sqrt{n}} =\frac{\sqrt\pi}{4 n^{3/2}}.$$

Answer 2

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}