Improper integral of $\int_{0}^{\infty}\dfrac{t}{e^{2\pi t}-1}dt$ I'm having trouble calculating $$\int_{0}^{\infty}{\dfrac{t}{e^{2\pi t}-1}dt}$$ I tried with substitution but I get to a point where I must calculate $$\int_{1}^{\infty}{\dfrac{\ln(y)}{y^2-y}}dy$$ and I still having the same problem.
Any ideas? 
 A: Redacting @achille hui's hint :
For $a>0\ ,\displaystyle \int_0^\infty te^{-ant}dt=\underbrace{\bigg[\frac{-te^{-ant}}{an}\bigg]_0^\infty}_{0}+\int_0^\infty\frac{e^{-ant}}{an}dt=\bigg[\frac{-e^{-ant}}{a^2n^2}\bigg]_0^\infty=\frac 1{a^2n^2}$
$\displaystyle f_N(t)=\sum\limits_{n=1}^N te^{-ant}=te^{-at}\frac{1-e^{-a(N+1)t}}{1-e^{-at}} \to\frac{t}{e^{at}-1}$ when $N\to\infty$ for any $t>0$.
Since $te^{-ant}\ge 0$ then $f_N\nearrow$ and we can apply monotone convergence theorem.
$\displaystyle \int_0^\infty \frac {tdt}{e^{at}-1}=\sum\limits_{n=1}^\infty \frac 1{a^2n^2}=\frac{\pi^2}{6a^2}$
A: If you expand $\frac{1}{e^{2\pi t}-1}$ as a geometric series, you may check that:
$$ I = \int_{0}^{+\infty}\frac{t\,dt}{e^{2\pi t}-1} =\sum_{n\geq 1}\int_{0}^{+\infty}t e^{-2n\pi t}\,dt=\sum_{n\geq 1}\frac{1}{4\pi^2 n^2}\int_{0}^{+\infty}s e^{-s}\,ds$$
hence:
$$ I = \frac{1}{4\pi^2}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{4\pi^2} = \color{red}{\frac{1}{24}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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It's amusing to recognize this integral as an application of
  Abel-Plana Formula:

\begin{align}
&\lim_{N \to \infty}\pars{\sum_{n = 0}^{N}n - \int_{0}^{N}t\,\dd t - {1 \over 2}\, \times 0 - {1 \over 2}\,N} = -
2\int_{0}^{\infty}{\Im\pars{0 + \ic t} \over \expo{2\pi t} - 1}\,\dd t +
{1 \over 2}\,B_{2}
\end{align}
where $\ds{B_{s}}$ is a Bernoulli Number. Note that $\ds{B_{2} = 1/6}$.

$$
\int_{0}^{\infty}{t \over \expo{2\pi t} - 1}\,\dd t =
{1 \over 4}\,B_{2} = {1 \over 4}\times{1 \over 6} = \bbx{1 \over 24}
$$
