How to calculate $ \int \limits_{0}^{\infty} \frac{ t^2 e^{-t}}{(1 + e^{-t})^2} dt $?

Question

I am trying to calculate

$$ \int \limits_{0}^{\infty} \frac{ t^2 e^{-t}}{(1 + e^{-t})^2} dt $$

I used variable replacement $u = e^{-t} $ and got the integral

$$ \int \limits_{0}^{1} \frac{ \ln^2u}{(1 + u)^2} du $$

Wolfram says that it is $\frac{\pi}{6}$. But I have not idea. How can I get this result?

Thank you in advance for any help!

Title and first integral in post are different! – Fakemistake Mar 20 '21 at 13:14 — Fakemistake, Mar 20 '21 at 13:14
I would assume the substitution was $u=e^{-t}$. – user Mar 20 '21 at 13:18 — user, Mar 20 '21 at 13:18
@Fakemistake, sorry for this! fixed – GThompson Mar 20 '21 at 13:20 — GThompson, Mar 20 '21 at 13:20

score 1 · Accepted Answer · answered Mar 20 '21 at 13:17

1

Note

\begin{align} \int_0^1\frac{\ln^2 u}{(u+1)^2}du &= \int_0^1{\ln^2 u}\>d(\frac{u}{u+1})\\ &=- 2\int_0^1\frac{\ln u}{u+1}du = -2 (-\frac{\pi^2}{12})=\frac{\pi^2}6 \end{align}

where $\int_0^1 \frac{\ln u}{u+1}dt =-\frac{\pi^2}{12}$

answered Mar 20 '21 at 13:17

Quanto

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How to calculate $ \int \limits_{0}^{\infty} \frac{ t^2 e^{-t}}{(1 + e^{-t})^2} dt $?

1 Answers1