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

  • 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

  • @Darshan Patil Can you help me understand why differentiating doesn't work just the way you solved https://math.stackexchange.com/questions/4292020/how-to-evaluate-the-integral-i-int-o-infty-fracx-sqrte2-pi-sqrtx/4292035#4292035 –  Nov 02 '21 at 18:14
  • @Parvathi Ayyangar, the differentiation do works. – RAHUL Nov 02 '21 at 18:18
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    The Laplace Transform $\int_0^\infty \frac{e^{-st}}{e^t-1},dt$ fails to exist and hence discussion of its derivative is meaningless. – Mark Viola Nov 02 '21 at 18:19
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    Laplace transform will work, if you regularize the integrand at $t=0$. Take, for instance, $I(s)=\int_0^\infty \frac{1-e^{-st}}{e^t-1}dt$. This integral converges, and $\int_0^\infty \frac{t^n}{e^t-1}dt=(-1)^{n+1}\frac{\partial ^n}{\partial s^n}I(s)\Big|{s=0}$. Making change $x=e^{-t}$ $$I(s)=\int_0^\infty\frac{1-t^s}{1-t}dt=-\sum{k=1}^\infty\Big(\frac{1}{k+s}-\frac{1}{k}\Big)=\gamma+\psi(1+s)$$ where $\psi(x)$ - digamma function. For $n>0$ $$\int_0^\infty \frac{t^n}{e^t-1}dt=(-1)^{n+1}\frac{\partial ^n}{\partial s^n}\psi(s+1)\Big|_{s=0}=(-1)^{n+1}\psi^{(n)}(1)=n!\zeta(n+1)$$ – Svyatoslav Nov 04 '21 at 05:46

3 Answers3

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

$$\begin{align*} I &= \int_0^{\infty}\frac t{e^t-1}dt\\ & \text{We substitute, } e^{-t} = x \implies t = -\ln x \implies dt = -\frac{dx}{x} \\ & \color{blue}{I = -\int_0^1 \frac {\ln x}{1-x}dx}\\ \end{align*}$$

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

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    Woww! You made that this is more than enough! –  Nov 02 '21 at 18:30
  • Thanks! This is what exactly I was looking you explained why differentiating was failing. –  Nov 02 '21 at 18:31
  • You mentioned polygamma function but I didn't find polygamma function in your answer? I would like to know –  Nov 02 '21 at 19:02
  • @ParvathiAyyangar Yes, I didn't mention but you can read on it Hint $\psi_1(1)$ $= \zeta(2) = \frac {\pi^2}{6}$ –  Nov 02 '21 at 19:04
  • for the first laplace transform Identity!
    • –  Nov 06 '21 at 03:57