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

Solution 1:

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.

• 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 $\Gamma(0)$

• 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 $$\begin{align*} F(s) = \mathcal{L}f(t) &= \int_0^{\infty}e^{-st}f(t)dt\\ & \implies \sum_{s=1}^{\infty}F(s) = \int_0^{\infty} \frac {f(t)}{e^t-1}dt\\ & \text{We can put } f(t) = t \implies \mathcal{L}(t) = \frac 1{s^2}\\ & \text{Thus,}\\ & \color{green}{I = \int_0^{\infty} \frac{t}{e^t-1}dt = \sum_{s=1}^{\infty}\frac {1}{s^2}=\zeta(2) = \frac {\pi^2}{6}}\\ \end{align*}$$

• $2]$ 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 $\psi_1(z)$

$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = -\int_0^1 \frac {x^{z-1}\ln x}{1-x}dx$$ $$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$

• $3]$ Using the integral form of even values of zeta function

$$\left|\zeta(2n) = \frac 1{(2n-1)!}\int_0^{\infty}\frac{x^{2n-1}}{e^x-1}dx\right|_{n =1}$$

• $4]$ More general form of the above integral in terms of Bernoulli's Number

$$\left|\int_0^{\infty} \frac {x^{2n-1}}{e^{px}-1}dx = (-1)^{n-1}\left(\frac {2\pi}{p}\right)^{2n} \frac{B_{2n}}{4n}\right|_{n, p = 1, 1}$$ $B_2 = \frac 1{6}$

• 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 $f_c(u = 0.0001) = 1.644934043288231 ≈ \pi^2/6$

Solution 2:

$$\begin{align}\int_0^{\infty} \frac{t}{e^t-1} dt &= \int_0^{\infty} \frac{te^{-t}}{1-e^{-t}}\,dt\\\\ &=\int_0^{\infty}t\sum_{n=1}^{\infty}e^{-nt} dt\\\\ &=\sum_{n=1}^{\infty}\int_0^{\infty}te^{-nt} \,dt\\\\ &=\sum_{n=1}^{\infty}\left.\left(-\frac{e^{-nt}(nt+1)}{n^2}\right)\right|_0^\infty \\\\ &=\sum_{n=1}^{\infty}\frac{1}{n^2}\\\\&=\frac{\pi^2}{6} \end{align}$$

The last sum is well known as the Basel problem and I'll let you fill in the details of interchanging summation and integration.