Question Number 211531 by MrGaster last updated on 12/Sep/24
$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{{x}}\right)^{\mathrm{2}} }\boldsymbol{{dx}}. \\ $$
Answered by Frix last updated on 13/Sep/24
![∫_0 ^∞ ((x/(sinh x)))^2 dx =^([t=(1/(e^(2x) −1))]) = (1/2)∫_0 ^∞ (ln ((t+1)/t))^2 dt =^([by parts]) =(1/2)[t(ln ((t+1)/t))^2 ]_0 ^∞ _( _(=0) ) +∫_0 ^∞ (((ln (t+1))/(t+1))−((ln t)/(t+1)))dt= =∫_0 ^∞ (((ln (t+1))/(t+1))−((ln t)/(t+1)))dt=(π^2 /6)](https://www.tinkutara.com/question/Q211559.png)
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{{x}}{\mathrm{sinh}\:{x}}\right)^{\mathrm{2}} {dx}\:\overset{\left[{t}=\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}{x}} −\mathrm{1}}\right]} {=} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}}\right)^{\mathrm{2}} {dt}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=\underset{\underset{=\mathrm{0}} {\underbrace{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}} {\frac{\mathrm{1}}{\mathrm{2}}\left[{t}\left(\mathrm{ln}\:\frac{{t}+\mathrm{1}}{{t}}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\infty} }+\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}+\mathrm{1}}−\frac{\mathrm{ln}\:{t}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}+\mathrm{1}}−\frac{\mathrm{ln}\:{t}}{{t}+\mathrm{1}}\right){dt}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$