Question Number 163842 by mnjuly1970 last updated on 11/Jan/22
$$ \\ $$$$\:\:\:\:\:\:\:{calculate} \\ $$$$\:\:\:\: \\ $$$$\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\:\:\mathscr{A}{rctanh}\:\left({x}\right)}{{x}^{\:} }\right)^{\:\mathrm{2}} \:{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:−−\:{m}.{n}\:−− \\ $$
Answered by amin96 last updated on 11/Jan/22
![𝛀=(1/4)∫_0 ^1 (1/x^2 )ln^2 (((1+x)/(1−x)))dx =^(((1−x)/(1+x))=t) (1/2)∫_0 ^1 ((ln^2 (t))/((1−t)^2 ))=^(IBP) =^(IBP) (1/2)[(t/(1−t))ln^2 t]_0 ^1 −(1/2)∫_0 ^1 (t/(1−t))×((2lnt)/t)dt=−∫_0 ^1 ((lnt)/(1−t))dt=^(t=1−t) =^(t=1−t) −∫_0 ^1 ((ln(1−t))/t)dx=Σ_(n=1) ^∞ (1/n)∫_0 ^1 t^(n−1) dt=Σ_(n=1) ^∞ (1/n^2 )=𝛇(2)](https://www.tinkutara.com/question/Q163847.png)
$$\boldsymbol{\Omega}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\frac{\mathrm{1}+\boldsymbol{\mathrm{x}}}{\mathrm{1}−\boldsymbol{\mathrm{x}}}\right)\boldsymbol{\mathrm{dx}}\:\overset{\frac{\mathrm{1}−\boldsymbol{\mathrm{x}}}{\mathrm{1}+\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{t}}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{t}}\right)}{\left(\mathrm{1}−\boldsymbol{\mathrm{t}}\right)^{\mathrm{2}} }\overset{\boldsymbol{\mathrm{IBP}}} {=} \\ $$$$\overset{\boldsymbol{\mathrm{IBP}}} {=}\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\boldsymbol{\mathrm{t}}}{\mathrm{1}−\boldsymbol{\mathrm{t}}}\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \boldsymbol{\mathrm{t}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{t}}}{\mathrm{1}−\boldsymbol{\mathrm{t}}}×\frac{\mathrm{2}\boldsymbol{\mathrm{lnt}}}{\boldsymbol{\mathrm{t}}}\boldsymbol{\mathrm{dt}}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{lnt}}}{\mathrm{1}−\boldsymbol{\mathrm{t}}}\boldsymbol{\mathrm{dt}}\overset{\boldsymbol{\mathrm{t}}=\mathrm{1}−\boldsymbol{\mathrm{t}}} {=} \\ $$$$\overset{\boldsymbol{\mathrm{t}}=\mathrm{1}−\boldsymbol{\mathrm{t}}} {=}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}−\boldsymbol{\mathrm{t}}\right)}{\boldsymbol{\mathrm{t}}}\boldsymbol{\mathrm{dx}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \boldsymbol{\mathrm{dt}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }=\boldsymbol{\zeta}\left(\mathrm{2}\right) \\ $$
Commented by mnjuly1970 last updated on 11/Jan/22
$${thanks}\:{alot}\:{sir}\:{amin}\:… \\ $$$${have}\:{a}\:{nice}\:{time} \\ $$