Question Number 33983 by abdo imad last updated on 28/Apr/18
$${find}\:\int_{\mathrm{1}\left(\right.} ^{\infty} \frac{\mathrm{1}}{{x}}{ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right){dx}. \\ $$
Commented by abdo mathsup 649 cc last updated on 03/May/18
$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{1}}{{x}}{ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right){dx}\:\:{changement} \\ $$$${x}=\frac{\mathrm{1}}{{t}}\:{give}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\:{ln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{{t}}{ln}\left(\:\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right){dt}\:\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:{this}\:{result}\:{is}\:{proved}\:. \\ $$