Question Number 150992 by talminator2856791 last updated on 17/Aug/21
$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{{x}}{{e}^{{x}} −\mathrm{1}}\:{dx}\:=\:? \\ $$$$\: \\ $$$$\: \\ $$
Commented by puissant last updated on 17/Aug/21
$$=\int_{\mathrm{0}} ^{\infty} \frac{{xe}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\bigstar \\ $$
Answered by Ar Brandon last updated on 17/Aug/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}}{{e}^{{x}} −\mathrm{1}}{dx}=\Gamma\left(\mathrm{2}\right)\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$
Answered by qaz last updated on 17/Aug/21
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}}{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}\mathrm{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{xe}^{−\mathrm{x}} }{\mathrm{1}−\mathrm{e}^{−\mathrm{x}} }\mathrm{dx}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\infty} \mathrm{xe}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} \mathrm{dx}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$