Evaluating Definite Integral $\int_0^\infty\frac{x}{e^x-1}dx$

Question

I am looking for alternative ways to solve the Basel problem using only real analysis and without using the infinite product for sine. I have transformed the summation into the given integral below. It looks like a deceivingly simply integral to evaluate, but none of the websites that I have plugged it into have been able to solve it. One of them even claimed that it was a divergent integral. Through numerical integration I have verified that the integral is equal to $\frac{\pi^2}{6}$. I am hoping that someone will be able to prove the integral is equal to its known closed form. For the purposes of this proof, start with the integral, and do not revert to the initial summation. $$\sum_{n=1}^\infty \frac{1}{n^2} = \int_0^\infty \frac x {e^x-1} \, dx = \frac{\pi^2} 6$$

$$\text{Transform Explanation:}$$

$$a(x)=\sum_{n=1}^\infty \frac{x^n}{n^2}$$ $$a(0)=0 \space\space\space\space\space\space\space\space a(1)=\sum_{n=1}^\infty \frac 1 {n^2}$$ $$a(1)=\int_0^1{a'(x)dx}$$ $$a'(x)=\frac{1}{x}{\sum_{n=1}^{\infty}{\frac{x^n}{n}}}$$ $$b(x)={\sum_{n=1}^{\infty}{\frac{x^n}{n}}} \space\space\space\space\space\space\space\space a'(x)=\frac{b(x)}{x}$$ $$b(0)=0 \space\space\space\space\space\space\space\space b(x)=\int_0^x{b'(t)dt}$$ $$b'(x)={\sum_{n=0}^{\infty}{x^n}}=\frac{1}{1-x} \space\space\space\space\space For \space\space -1<x<1$$ $$b(x)=\int_0^x{\frac{dt}{1-t}}=-\ln|1-x|$$ $$a'(x)=\frac{-\ln|1-x|}{x}$$ $$a(1)=\int_0^1{\frac{-\ln|1-x|}{x}dx}=\int_0^1 \frac{-\ln(1-x)} x \, dx$$ $u=-\ln(1-x) \space\space\space\space -u=\ln(1-x) \space\space\space\space e^{-u} = 1-x \space\space\space\space x=1-e^{-u} \space\space\space\space dx=e^{-u} \, du$ $u(x)=-\ln(1-x) \space\space\space\space u(0^+)=0 \space\space\space\space u(1^-)=\infty$ $$\int_0^1 \frac{-\ln(1-x)}{x} \, dx =\int_0^\infty \frac{ue^{-u}}{1-e^{-u}} \, du$$ $$a(1)=\int_0^\infty\frac{x}{e^x-1}\,dx=\sum_{n=1}^\infty \frac 1 {n^2}$$ Thank you for your time!

Answer 1

Note that by enforcing the substitution $x\to -\log(1-x)$, we can write

$$\begin{align} \int_0^\infty \frac{x}{e^x-1}\,dx&=-\int_0^1\frac{\log(1-x)}{x}\,dx\\\\ &=\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy\tag 1 \end{align}$$

Then, in THIS ANSWER, I used the transformation $x=s+t$ and $y=s-t$ to directly evaluate the transformed integral

$$\begin{align} \int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\ &=2\frac{\pi^2}{6^2}+\frac{\pi^2}{9}\\\\ &=\frac{\pi^2}{6} \end{align}$$

as expected!

Answer 2

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With Abel-Plana Formula ("$\ds{\mathsf{\underline{finite\ version}}}$") :

\begin{align} \sum_{x = 0}^{n}x & = \int_{0}^{n}x\,\dd x + {1 \over 2}\,0 + {1 \over 2}\,n - 2\,\Im\int_{0}^{\infty}{\ic x \over \expo{2\pi x} - 1}\,\dd x + {B_{2} \over 2} \end{align}

$\ds{B_{s}}$ is a Bernoulli Number. Note that $\ds{B_{2} = {1 \over 6}}$.

Then, \begin{align} {n\pars{n + 1} \over 2} & = {n^{2} \over 2} + {n \over 2} - 2\,{1 \over \pars{2\pi}^{2}}\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x + {1 \over 12} \\[5mm] \implies & \bbx{\int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x = {1 \over 2}\,\pars{2\pi}^{2}\,{1 \over 12} = {\pi^{2} \over 6}} \end{align}