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Question Number 137799 by mnjuly1970 last updated on 06/Apr/21

..... mathematical .. ... ... analysis.... evaluate :: 𝛗=∫_0 ^( 1) (((ln^2 (1βˆ’x^2 ))/x))=?

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.....\:{mathematical}\:..\:...\:...\:{analysis}.... \\ $$$$\:\:\:\:\:\:\:{evaluate}\:::\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{{x}}\right)=? \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 06/Apr/21

∫_0 ^1 ((log^2 (1βˆ’x^2 ))/x)dx =(1/2)∫_0 ^1 ((log^2 (1βˆ’u))/u)du=[(1/2)log^2 (1βˆ’u)log(u)]_0 ^1 +∫_0 ^1 log(1βˆ’u)((log(u))/(1βˆ’u)) =∫_0 ^1 ((log(u)log(1βˆ’u))/u)du =[βˆ’log(u)Ξ£_(n=1) ^∞ (x^n /n^2 )]_0 ^1 +Ξ£_(n=1) ^∞ ∫_0 ^1 (x^(nβˆ’1) /n^2 )dx =Ξ£_(n=1) ^∞ (1/n^3 )=ΞΆ(3)=1.20206...

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}^{\mathrm{2}} \left(\mathrm{1}βˆ’{u}\right)}{{u}}{du}=\left[\frac{\mathrm{1}}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{1}βˆ’{u}\right){log}\left({u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}βˆ’{u}\right)\frac{{log}\left({u}\right)}{\mathrm{1}βˆ’{u}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left({u}\right){log}\left(\mathrm{1}βˆ’{u}\right)}{{u}}{du} \\ $$$$=\left[βˆ’{log}\left({u}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} +\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}βˆ’\mathrm{1}} }{{n}^{\mathrm{2}} }{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right)=\mathrm{1}.\mathrm{20206}... \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 06/Apr/21

thanks alot mr payan very nice ....

$${thanks}\:{alot}\:{mr}\:{payan}\: \\ $$$$\:{very}\:{nice}\:.... \\ $$

Answered by EnterUsername last updated on 06/Apr/21

∫_0 ^1 ((ln^2 (1βˆ’x^2 ))/x)dx=(1/2)∫_0 ^1 ((ln^2 (1βˆ’u))/u)du=(1/2)∫_0 ^1 ((ln^2 u)/(1βˆ’u))du f(s)=(1/2)∫_0 ^1 ((u^(sβˆ’1) ln^2 u)/(1βˆ’u))du=βˆ’(1/2)Οˆβ€²β€²(s) Ο†=f(1)=βˆ’(1/2)Οˆβ€²β€²(1)=(1/2)Ξ£_(n=0) ^∞ (2/((n+1)^3 ))=Ξ£_(n=1) ^∞ (1/n^3 )=ΞΆ(3)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{x}^{\mathrm{2}} \right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}βˆ’{u}\right)}{{u}}{du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {u}}{\mathrm{1}βˆ’{u}}{du} \\ $$$${f}\left({s}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{s}βˆ’\mathrm{1}} {ln}^{\mathrm{2}} {u}}{\mathrm{1}βˆ’{u}}{du}=βˆ’\frac{\mathrm{1}}{\mathrm{2}}\psi''\left({s}\right) \\ $$$$\phi={f}\left(\mathrm{1}\right)=βˆ’\frac{\mathrm{1}}{\mathrm{2}}\psi''\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\zeta\left(\mathrm{3}\right) \\ $$

Commented by mnjuly1970 last updated on 06/Apr/21

grateful ..thank you very much..

$$\:\:{grateful}\:..{thank}\:{you}\:{very}\:{much}.. \\ $$

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