Question Number 133857 by mnjuly1970 last updated on 24/Feb/21
![.....#advanced ............... calculus#..... prove that ::: π=β«_0 ^( 1) ((ln^2 (1βx))/x)dx=^? 2ΞΆ(3) =^(1βx=t) β«_0 ^( 1) ((ln^2 (t))/(1βt))dt=β«_0 ^( 1) Ξ£_(n=0) ^β ln^2 (t).t^n dt =Ξ£_(n=0) ^β {[(t^(n+1) /(n+1))ln^2 (t)]_0 ^1 β(2/(n+1))β«_0 ^( 1) t^n ln(t) =β2Ξ£_(n=0) ^β (1/(n+1)){[(t^(n+1) /(n+1))ln(t)]_0 ^1 β(1/(n+1))β«_0 ^( 1) t^n dt} =2Ξ£_(n=0) ^β (1/((n+1)^3 ))=2Ξ£_(n=1) ^β (1/n^3 )=2ΞΆ(3) .................... π=2ΞΆ(3) .................... ..........m.n.july.1970.........](https://www.tinkutara.com/question/Q133857.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..#{advanced}\:\:\:\:……………\:\:\:{calculus}#….. \\ $$$$\:\:\:\:{prove}\:\:{that}\::::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}\right)}{{x}}{dx}\overset{?} {=}\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{1}β{x}={t}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}β{t}}{dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{ln}^{\mathrm{2}} \left({t}\right).{t}^{{n}} {dt} \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}^{\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} β\frac{\mathrm{2}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} {ln}\left({t}\right)\right. \\ $$$$\:\:\:\:=β\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\left\{\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} β\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} {dt}\right\} \\ $$$$\:\:\:\:=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………..\:\:\:\boldsymbol{\phi}=\mathrm{2}\zeta\left(\mathrm{3}\right)\:……………….. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….{m}.{n}.{july}.\mathrm{1970}……… \\ $$
Answered by Dwaipayan Shikari last updated on 24/Feb/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left({t}\right){t}^{{n}} {dt} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} {u}^{\mathrm{2}} {e}^{β{u}} {dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{3}} }=\mathrm{2}\zeta\left(\mathrm{3}\right)\:\:\:\:\:\:\:{u}=β{logt} \\ $$
Commented by mnjuly1970 last updated on 24/Feb/21
$${nice}\:\:{solution}.. \\ $$
Answered by ΓΓ―= last updated on 04/Mar/21
$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}\right)}{{x}}{dx}={lnxln}^{\mathrm{2}} \left(\mathrm{1}β{x}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnx}}{\mathrm{1}β{x}}\centerdot\mathrm{2}{ln}\left(\mathrm{1}β{x}\right){dx} \\ $$$$\mathrm{1}β{x}={t} \\ $$$$\phi=β\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{ln}\left(\mathrm{1}β{t}\right)}{{t}}\centerdot\mathrm{2}{lntdt}={Li}_{\mathrm{2}} \left({t}\right)\mathrm{2}{lnt}\mid_{\mathrm{1}} ^{\mathrm{0}} β\mathrm{2}\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{Li}_{\mathrm{2}} \left({t}\right)}{{t}}{dt} \\ $$$$=β\mathrm{2}{Li}_{\mathrm{3}} \left({t}\right)\mid_{\mathrm{1}} ^{\mathrm{0}} =\mathrm{2}{Li}_{\mathrm{3}} \left(\mathrm{1}\right)=\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$