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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::Ο•=∫01(ln2(1βˆ’x2)x)=?
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...
∫01log2(1βˆ’x2)xdx=12∫01log2(1βˆ’u)udu=[12log2(1βˆ’u)log(u)]01+∫01log(1βˆ’u)log(u)1βˆ’u=∫01log(u)log(1βˆ’u)udu=[βˆ’log(u)βˆ‘βˆžn=1xnn2]01+βˆ‘βˆžn=1∫01xnβˆ’1n2dx=βˆ‘βˆžn=11n3=ΞΆ(3)=1.20206…
Commented by mnjuly1970 last updated on 06/Apr/21
thanks alot mr payan    very nice ....
thanksalotmrpayanverynice….
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)
∫01ln2(1βˆ’x2)xdx=12∫01ln2(1βˆ’u)udu=12∫01ln2u1βˆ’uduf(s)=12∫01usβˆ’1ln2u1βˆ’udu=βˆ’12Οˆβ€³(s)Ο•=f(1)=βˆ’12Οˆβ€³(1)=12βˆ‘βˆžn=02(n+1)3=βˆ‘βˆžn=11n3=ΞΆ(3)
Commented by mnjuly1970 last updated on 06/Apr/21
  grateful ..thank you very much..
grateful..thankyouverymuch..

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