In an attempt to find $I = \int_0^\infty \frac{t}{e^t-1}dt$ I was trying to solve $I = \int_0^\infty \frac{t}{e^t-1}dt$

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*My approach

I took the more general form of integral $f(s) = \int_0^{\infty}\frac{e^{-st}}{e^t-1}dt$ the same way as How to evaluate the integral $I = \int_o^{\infty} \frac{x}{\sqrt{e^{2\pi\sqrt{x}}-1}}dx$?
So, my answer $$I = -\left[\frac{\partial f(s)}{\partial s}\right]_{s =0}$$ but that simply doesn't work.
However, $\int_0^\infty \frac{t}{e^t-1}dt$ without using series doesn't answer my question as I'm interested in finding why finding derivative at $s = 0$ doesn't work
Also, I'm interested in finding a number of different ways I can solve this problem
 A: Well! This integral is really interesting and has a lot to do with Bernoulli's Numbers, Riemann Zeta function, Polygamma function, in the last what you were trying is nothing less than Laplace Transformation
$$I = \int_0^{\infty}\frac {t}{e^t-1}dt$$
First I'll try to answer why $$I =-\frac{\partial}{\partial s}\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$$ is failing to conclude answer and after that, I'll try to add few other ways you can conclude to the answer.

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*Why $I =-\frac{\partial}{\partial s}\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$ fails
Answer On the very first look you can conclude that the definite integral $\int_0^{\infty}\frac {e^{-st}}{e^t-1}dt$ doesn't exist(Undefined)
Hint Lower limit of definite integral is $0$ and when $t =0 $ numerator = $1$ but denominator $=0$ However, for $\int_0^{\infty}\frac {t}{e^t-1}dt$ it's totally different
Even if you try to find by taking laplace transformation you'll eventually end up for $\Gamma(0)$

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*How to solve using Transformation only?
This may answer your question but as you asked for a few other ways to solve this integral I'm adding a few transformation based answers.


*$1$] Using Laplace Transform Identity
I'm unable to provide a link from where I found this but if I could remember you can find it in "Handbook of Mathematics for Engineer's & Scientists" by Andrei & Alexander
If
$$\begin{align*}
F(s) = \mathcal{L}f(t) 
&= \int_0^{\infty}e^{-st}f(t)dt\\
& \implies \sum_{s=1}^{\infty}F(s) = \int_0^{\infty} \frac {f(t)}{e^t-1}dt\\
& \text{We can put } f(t) = t \implies \mathcal{L}(t) = \frac 1{s^2}\\
& \text{Thus,}\\
& \color{green}{I = \int_0^{\infty} \frac{t}{e^t-1}dt = \sum_{s=1}^{\infty}\frac {1}{s^2}=\zeta(2) = \frac {\pi^2}{6}}\\
\end{align*}$$

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*$2]$ Using diagamma function
$$\begin{align*}
I 
&= \int_0^{\infty}\frac t{e^t-1}dt\\
& \text{We substitute, } e^{-t} = x \implies t = -\ln x \implies dt = -\frac{dx}{x} \\ 
& \color{blue}{I = -\int_0^1 \frac {\ln x}{1-x}dx}\\
\end{align*}$$
Now,
$$\color{red}{\psi_0 (z) = -\gamma + \int_0^{1} \frac {1-x^{z-1}}{1-x}dx}$$
We'll differentiate and that is trigamma function $\psi_1(z)$
$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = -\int_0^1 \frac {x^{z-1}\ln x}{1-x}dx$$
$$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$

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*$3]$ Using the integral form of even values of zeta function
$$\left|\zeta(2n) = \frac 1{(2n-1)!}\int_0^{\infty}\frac{x^{2n-1}}{e^x-1}dx\right|_{n =1}$$

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*$4]$ More general form of the above integral in terms of Bernoulli's Number
$$\left|\int_0^{\infty} \frac {x^{2n-1}}{e^{px}-1}dx = (-1)^{n-1}\left(\frac {2\pi}{p}\right)^{2n} \frac{B_{2n}}{4n}\right|_{n, p = 1, 1}$$
$B_2 = \frac 1{6}$

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*My last attempt: The cosine Transformation
$$f_c(u) = \int_0^{\infty}f(x)\cos(ux)dx$$
for $$\text{if  }f(x) = \frac x{e^{ax}-1} \implies f_c(u) = \frac 1{2u^2} - \frac {\pi^2}{2a^2\sinh^2(\pi a^{-1}u)} $$
I tried to take the limit as $u \to 0$ but didn't work maybe because cosine transformation is defined for $u \in (0, \infty)$ but at $u = 0.0001$ it's $f_c(u = 0.0001) = 1.644934043288231 ≈ \pi^2/6$
A: $$\begin{align}\int_0^{\infty} \frac{t}{e^t-1} dt &= \int_0^{\infty} \frac{te^{-t}}{1-e^{-t}}\,dt\\\\
&=\int_0^{\infty}t\sum_{n=1}^{\infty}e^{-nt}  dt\\\\
&=\sum_{n=1}^{\infty}\int_0^{\infty}te^{-nt}  \,dt\\\\
&=\sum_{n=1}^{\infty}\left.\left(-\frac{e^{-nt}(nt+1)}{n^2}\right)\right|_0^\infty \\\\
&=\sum_{n=1}^{\infty}\frac{1}{n^2}\\\\&=\frac{\pi^2}{6}
\end{align}$$
The last sum is well known as the Basel problem and I'll let you fill in the details of interchanging summation and integration.
A: $$\int\limits_0^\infty \dfrac{t^{2-1}}{e^t-1}\,\text dt = \Gamma(2)\zeta(2)
=\dfrac{\pi^2}6$$
(Riemann zeta function)
