Question Number 166180 by mnjuly1970 last updated on 15/Feb/22
![prove that π=β«_0 ^( 1) (( ln^( 2) (1βx ))/x^( 2) ) dx = 2 ΞΆ (2) βββproofβββ π= [((β1)/x) ln^( 2) (1βx) ]_0 ^1 ββ«_0 ^( 1) ((2ln(1βx))/(x(1βx)))dx =βlim_( ΞΎβ1^β ) (1/ΞΎ)ln^( 2) (1βΞΎ)β2{ β«_0 ^( 1) ((ln(1βx))/(1βx))dx+β«_0 ^( 1) ((ln(1βx))/x)dx} = βlim_( ΞΎβ1^β ) {(1/ΞΎ)ln^( 2) (1βΞΎ)+ln^( 2) (1 βΞΎ)}+2 ΞΆ(2) =lim_(ΞΎβ1^β ) (((ΞΎβ1)/ΞΎ))ln^( 2) (1βΞΎ) +2ΞΆ(2) =_(ΞΎβ1^β , Ξ΄β0^( +) ) ^(1βΞΎ= Ξ΄) [lim_( Ξ΄β0^( +) ) (((βΞ΄)/(1βΞ΄)))ln^2 (Ξ΄)=0] +2ΞΆ(2) β m.n β΄ π = 2 ΞΆ(2)](https://www.tinkutara.com/question/Q166180.png)
$$ \\ $$$$\:\:\:\:\:\:{prove}\:\:{that} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}β{x}\:\right)}{{x}^{\:\mathrm{2}} }\:{dx}\:=\:\mathrm{2}\:\zeta\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:βββ{proof}βββ \\ $$$$\:\:\:\:\boldsymbol{\phi}=\:\left[\frac{β\mathrm{1}}{{x}}\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}β{x}\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} β\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{2}{ln}\left(\mathrm{1}β{x}\right)}{{x}\left(\mathrm{1}β{x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:=β{lim}_{\:\xi\rightarrow\mathrm{1}^{β} } \frac{\mathrm{1}}{\xi}{ln}^{\:\mathrm{2}} \left(\mathrm{1}β\xi\right)β\mathrm{2}\left\{\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}β{x}\right)}{\mathrm{1}β{x}}{dx}+\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}β{x}\right)}{{x}}{dx}\right\} \\ $$$$\:\:\:\:\:\:\:=\:β{lim}_{\:\xi\rightarrow\mathrm{1}^{β} } \left\{\frac{\mathrm{1}}{\xi}{ln}^{\:\mathrm{2}} \left(\mathrm{1}β\xi\right)+{ln}^{\:\mathrm{2}} \left(\mathrm{1}\:β\xi\right)\right\}+\mathrm{2}\:\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:={lim}_{\xi\rightarrow\mathrm{1}^{β} } \left(\frac{\xiβ\mathrm{1}}{\xi}\right){ln}^{\:\mathrm{2}} \left(\mathrm{1}β\xi\right)\:+\mathrm{2}\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\underset{\xi\rightarrow\mathrm{1}^{β} \:,\:\delta\rightarrow\mathrm{0}^{\:+} } {\overset{\mathrm{1}β\xi=\:\delta} {=}}\left[{lim}_{\:\delta\rightarrow\mathrm{0}^{\:+} } \left(\frac{β\delta}{\mathrm{1}β\delta}\right){ln}^{\mathrm{2}} \left(\delta\right)=\mathrm{0}\right]\:+\mathrm{2}\zeta\left(\mathrm{2}\right)\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\boldsymbol{\phi}\:=\:\mathrm{2}\:\zeta\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$