Question

Challenge Problem
Evaluate
\[\large \int\limits_{0}^{\infty} \ln(\frac{e^x+1}{e^x-1}) \ dx\]

Answer 1

seems a bit tough.. http://www.wolframalpha.com/input/?i=integrate+ln%28%28e%5Ex+%2B1%29%2F%28e%5E-x+-1%29%29

Answer 2

does a substitution e^u = e^x + 1 help? im super tired + bit hung over atm so expect errors here lol

\[\int\limits{\ln(e^x+1)}dx = \int\limits{\ln(e^u)\frac{e^u}{e^x}}du = \int\limits{\frac{ue^u}{e^u - 1}}du\]

Answer 3

then by parts probably
\[\int\limits{\frac{ue^u}{e^u - 1}}du = u \ln(e^u - 1) - \int\limits{\ln(e^u-1)}du\]

Answer 4

i think im going in circles to be honest..

Answer 5

but circles are wonderful shapes so im not unhappy

Answer 6

aw that sucks.. just looked at the wolfram lol

Answer 7

let
\[ u=\ln(\frac{e^x+1}{e^x-1}) \]
then
\[ dx=-\frac{2e^u}{e^{2u}-1}du\]
so u have
\[ I=\int_{\infty}^{0}-\frac{2ue^u}{e^{2u}-1}du=\int_{0}^{\infty}\frac{2ue^{-u}}{1-e^{-2u}}du\]
\[\frac{1}{1-e^{-2u}}=1+e^{-2u}+e^{-4u}+...=\sum_{n=0}^{\infty} e^{-2n u}\]
hence
\[I=\int_{0}^{\infty} 2ue^{-u} \sum_{n=0}^{\infty} e^{-2n u} du=2\sum_{n=0}^{\infty} \int_{0}^{\infty} ue^{-(2n+1)u}du=2\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=2\frac{\pi^2}{8}=\frac{\pi^2}{4}\]

Answer 8

is this helpful?\[\int\limits_{0}^{\infty} \ln\left(\frac{e^x+1}{e^x-1}\right) \ \text dx= \int\limits_{0}^{\infty} \ln({e^x+1)} \ \text dx-\int\limits_{0}^{\infty} \ln{(e^x-1)}\text dx\]