Question-141017

Question Number 141017 by 777316 last updated on 14/May/21

Commented by mnjuly1970 last updated on 15/May/21

$$\:\:{hello}\:{mr}\:: \\ $$$$\:\:{what}\:“{font}“\:{and}\:“\:{app}“\:{did}\:{you} \\ $$$$\:{use}\:{to}\:{write}\:{math}?? \\ $$$$\:\: \\ $$

Answered by mnjuly1970 last updated on 14/May/21

a_n : =^(x^n =t) ∫_0 ^( 1) ((ln(1−t))/t^(1/n) )((1/n)t^((1/n)−1) )dt =(1/n)∫_0 ^( 1) ((ln(1−t))/t)dt=((−li_2 (1))/n)=((−π^2 )/(6n)) 𝛗:=Σ_(n=1) ^∞ (a_n /n) =((−π^2 )/6)Σ_(n=1) ^∞ (1/n^2 ) =((−π^4 )/(36)) ...✓

$$\:\:\:\:\:\:{a}_{{n}} :\:\overset{{x}^{{n}} ={t}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}^{\frac{\mathrm{1}}{{n}}} }\left(\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=\frac{−{li}_{\mathrm{2}} \left(\mathrm{1}\right)}{{n}}=\frac{−\pi^{\mathrm{2}} }{\mathrm{6}{n}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}_{{n}} }{{n}}\:=\frac{−\pi^{\mathrm{2}} }{\mathrm{6}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{−\pi^{\mathrm{4}} }{\mathrm{36}}\:…\checkmark \\ $$

Answered by Dwaipayan Shikari last updated on 14/May/21

∫_0 ^1 ((log(1−x^n ))/x)dx x^n =u =(1/n)∫_0 ^1 ((log(1−u))/u)du=−(π^2 /(6n)) Σ_(n=1) ^∞ −(π^2 /(6n^2 ))=−(π^4 /(36))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{x}^{{n}} \right)}{{x}}{dx}\:\:\:{x}^{{n}} ={u} \\ $$$$=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{u}\right)}{{u}}{du}=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}{n}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{4}} }{\mathrm{36}} \\ $$

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