Question Number 137799 by mnjuly1970 last updated on 06/Apr/21
$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:…..\:{mathematical}\:..\:…\:…\:{analysis}…. \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{{x}}\right)=? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 06/Apr/21
![β«_0 ^1 ((log^2 (1βx^2 ))/x)dx =(1/2)β«_0 ^1 ((log^2 (1βu))/u)du=[(1/2)log^2 (1βu)log(u)]_0 ^1 +β«_0 ^1 log(1βu)((log(u))/(1βu)) =β«_0 ^1 ((log(u)log(1βu))/u)du =[βlog(u)Ξ£_(n=1) ^β (x^n /n^2 )]_0 ^1 +Ξ£_(n=1) ^β β«_0 ^1 (x^(nβ1) /n^2 )dx =Ξ£_(n=1) ^β (1/n^3 )=ΞΆ(3)=1.20206...](https://www.tinkutara.com/question/Q137801.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}β{u}\right)}{{u}}{du}=\left[\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{1}β{u}\right){log}\left({u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}β{u}\right)\frac{{log}\left({u}\right)}{\mathrm{1}β{u}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({u}\right){log}\left(\mathrm{1}β{u}\right)}{{u}}{du} \\ $$$$=\left[β{log}\left({u}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}β\mathrm{1}} }{{n}^{\mathrm{2}} }{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)=\mathrm{1}.\mathrm{20206}… \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 06/Apr/21
$${thanks}\:{alot}\:{mr}\:{payan}\: \\ $$$$\:{very}\:{nice}\:…. \\ $$
Answered by EnterUsername last updated on 06/Apr/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{x}^{\mathrm{2}} \right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}β{u}\right)}{{u}}{du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {u}}{\mathrm{1}β{u}}{du} \\ $$$${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{s}β\mathrm{1}} {ln}^{\mathrm{2}} {u}}{\mathrm{1}β{u}}{du}=β\frac{\mathrm{1}}{\mathrm{2}}\psi''\left({s}\right) \\ $$$$\phi={f}\left(\mathrm{1}\right)=β\frac{\mathrm{1}}{\mathrm{2}}\psi''\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right) \\ $$
Commented by mnjuly1970 last updated on 06/Apr/21
$$\:\:{grateful}\:..{thank}\:{you}\:{very}\:{much}.. \\ $$