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Question Number 133857 by mnjuly1970 last updated on 24/Feb/21
              .....#advanced    ...............   calculus#.....      prove  that :::  𝛗=∫_0 ^( 1) ((ln^2 (1βˆ’x))/x)dx=^? 2ΞΆ(3)          =^(1βˆ’x=t) ∫_0 ^( 1) ((ln^2 (t))/(1βˆ’t))dt=∫_0 ^( 1) Ξ£_(n=0) ^∞ ln^2 (t).t^n dt       =Ξ£_(n=0) ^∞ {[(t^(n+1) /(n+1))ln^2 (t)]_0 ^1 βˆ’(2/(n+1))∫_0 ^( 1) t^n ln(t)      =βˆ’2Ξ£_(n=0) ^∞ (1/(n+1)){[(t^(n+1) /(n+1))ln(t)]_0 ^1 βˆ’(1/(n+1))∫_0 ^( 1) t^n dt}      =2Ξ£_(n=0) ^∞ (1/((n+1)^3 ))=2Ξ£_(n=1) ^∞ (1/n^3 )=2ΞΆ(3)                              ....................   𝛗=2ΞΆ(3) ....................                              ..........m.n.july.1970.........
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..#{advanced}\:\:\:\:……………\:\:\:{calculus}#….. \\ $$$$\:\:\:\:{prove}\:\:{that}\::::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx}\overset{?} {=}\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{1}βˆ’{x}={t}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}βˆ’{t}}{dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{ln}^{\mathrm{2}} \left({t}\right).{t}^{{n}} {dt} \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{2}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} {ln}\left({t}\right)\right. \\ $$$$\:\:\:\:=βˆ’\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} {dt}\right\} \\ $$$$\:\:\:\:=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………..\:\:\:\boldsymbol{\phi}=\mathrm{2}\zeta\left(\mathrm{3}\right)\:……………….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{m}.{n}.{july}.\mathrm{1970}……… \\ $$
Answered by Dwaipayan Shikari last updated on 24/Feb/21
Ξ£_(n=1) ^∞ ∫_0 ^1 log^2 (t)t^n dt  =Ξ£_(n=1) ^∞ (1/n^3 )∫_0 ^∞ u^2 e^(βˆ’u) dt=Ξ£_(n=1) ^∞ (2/n^3 )=2ΞΆ(3)       u=βˆ’logt
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left({t}\right){t}^{{n}} {dt} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{2}} {e}^{βˆ’{u}} {dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{3}} }=\mathrm{2}\zeta\left(\mathrm{3}\right)\:\:\:\:\:\:\:{u}=βˆ’{logt} \\ $$
Commented by mnjuly1970 last updated on 24/Feb/21
nice  solution..
$${nice}\:\:{solution}.. \\ $$
Answered by Ñï= last updated on 04/Mar/21
Ο†=∫_0 ^1 ((ln^2 (1βˆ’x))/x)dx=lnxln^2 (1βˆ’x)∣_0 ^1 +∫_0 ^1 ((lnx)/(1βˆ’x))βˆ™2ln(1βˆ’x)dx  1βˆ’x=t  Ο†=βˆ’βˆ«_1 ^0 ((ln(1βˆ’t))/t)βˆ™2lntdt=Li_2 (t)2lnt∣_1 ^0 βˆ’2∫_1 ^0 ((Li_2 (t))/t)dt  =βˆ’2Li_3 (t)∣_1 ^0 =2Li_3 (1)=2ΞΆ(3)
$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}\right)}{{x}}{dx}={lnxln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}βˆ’{x}}\centerdot\mathrm{2}{ln}\left(\mathrm{1}βˆ’{x}\right){dx} \\ $$$$\mathrm{1}βˆ’{x}={t} \\ $$$$\phi=βˆ’\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{ln}\left(\mathrm{1}βˆ’{t}\right)}{{t}}\centerdot\mathrm{2}{lntdt}={Li}_{\mathrm{2}} \left({t}\right)\mathrm{2}{lnt}\mid_{\mathrm{1}} ^{\mathrm{0}} βˆ’\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt} \\ $$$$=βˆ’\mathrm{2}{Li}_{\mathrm{3}} \left({t}\right)\mid_{\mathrm{1}} ^{\mathrm{0}} =\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$

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