Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 137251 by mnjuly1970 last updated on 31/Mar/21

               ......Advanced  ...  calculus......      𝛗=∫_0 ^( 1) x^2 ln(x)ln(1βˆ’x)dx=???

......Advanced...calculus......Ο•=∫01x2ln(x)ln(1βˆ’x)dx=???

Answered by Ar Brandon last updated on 31/Mar/21

Ο†=∫_0 ^1 x^2 ln(x)ln(1βˆ’x)dx=(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) βˆ’βˆ«_0 ^1 x^(2+Ξ±) Ξ£_(n=1) ^∞ (x^n /n)dx     =(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) βˆ’Ξ£_(n=1) ^∞ (1/n)∫x^(2+Ξ±+n) dx=(βˆ‚/βˆ‚Ξ±)∣_(Ξ±=0) βˆ’Ξ£_(n=1) ^∞ (1/(n(3+Ξ±+n)))     =Ξ£_(n=1) ^∞ (1/(n(n+3)^2 ))=Ξ£_(n=1) ^∞ ((1/(9n))βˆ’(1/(9(n+3)))βˆ’(1/(3(n+3)^2 )))     =(1/9)Ξ£_(n=1) ^∞ ((1/n)βˆ’(1/(n+3)))βˆ’(1/3)(Ξ£_(n=1) ^∞ (1/n^2 )βˆ’(1+(1/2^2 )+(1/3^2 )))     =(1/9)(1+(1/2)+(1/3))βˆ’(1/3)(ΞΆ(2)βˆ’(1+(1/2^2 )+(1/3^2 )))     =((11)/(54))βˆ’(Ο€^2 /(18))+((49)/(108))=((71)/(108))βˆ’(Ο€^2 /(18))

Ο•=∫01x2ln(x)ln(1βˆ’x)dx=βˆ‚βˆ‚Ξ±βˆ£Ξ±=0βˆ’βˆ«01x2+Ξ±βˆ‘βˆžn=1xnndx=βˆ‚βˆ‚Ξ±βˆ£Ξ±=0βˆ’βˆ‘βˆžn=11n∫x2+Ξ±+ndx=βˆ‚βˆ‚Ξ±βˆ£Ξ±=0βˆ’βˆ‘βˆžn=11n(3+Ξ±+n)=βˆ‘βˆžn=11n(n+3)2=βˆ‘βˆžn=1(19nβˆ’19(n+3)βˆ’13(n+3)2)=19βˆ‘βˆžn=1(1nβˆ’1n+3)βˆ’13(βˆ‘βˆžn=11n2βˆ’(1+122+132))=19(1+12+13)βˆ’13(ΞΆ(2)βˆ’(1+122+132))=1154βˆ’Ο€218+49108=71108βˆ’Ο€218

Commented by Ar Brandon last updated on 31/Mar/21

:)

:)

Commented by Dwaipayan Shikari last updated on 31/Mar/21

Even shorter :)

Evenshorter:)

Commented by mnjuly1970 last updated on 31/Mar/21

thanks alot mr brandon...

thanksalotmrbrandon...

Commented by mnjuly1970 last updated on 31/Mar/21

grateful...

grateful...

Commented by Ar Brandon last updated on 31/Mar/21

Youβ€²re welcome, Sir !

Youβ€²rewelcome,Sir!

Commented by Dwaipayan Shikari last updated on 31/Mar/21

But , whenever you are calculating   ∫_0 ^1 x^Ο€ (1βˆ’x)^e log(x)log(1βˆ’x) dx  you need this Ξ½β€²(e+1,Ο€+1) ;βˆ’)

But,wheneveryouarecalculating∫01xΟ€(1βˆ’x)elog(x)log(1βˆ’x)dxyouneedthisΞ½β€²(e+1,Ο€+1);βˆ’)

Commented by Ar Brandon last updated on 31/Mar/21

OK, bro. Thanks πŸ˜ŠπŸ˜ƒ

OK, bro. Thanks πŸ˜ŠπŸ˜ƒ

Answered by Dwaipayan Shikari last updated on 31/Mar/21

∫_0 ^1 x^2 log(x)log(1βˆ’x)dx=Ξ½β€²(3,1)  ∫_0 ^1 x^(aβˆ’1) (1βˆ’x)^(bβˆ’1) dx=((Ξ“(a)Ξ“(b))/(Ξ“(a+b)))=Ξ½(a,b)  ∫_0 ^1 x^(aβˆ’1) (1βˆ’x)^(bβˆ’1) log(1βˆ’x)log(x)=Ξ½β€²(a,b)  =(βˆ‚/βˆ‚a)(((Ξ“(a)Ξ“β€²(1)βˆ’Οˆ(a+1)Ξ“(1)Ξ“(a))/(Ξ“(a+1))))  =(βˆ‚/βˆ‚a)(((βˆ’Ξ³)/a)βˆ’((ψ(a+1))/a))=(Ξ³/a^2 )+((ψ(1+a))/a^2 )βˆ’((Οˆβ€²(a+1))/a) ∣_(a=3)   =(Ξ³/9)+((ψ(4))/9)βˆ’((Οˆβ€²(4))/a)=((Ξ³βˆ’Ξ³+1+(1/2)+(1/3))/9)βˆ’(1/3)((Ο€^2 /6)βˆ’1βˆ’(1/2^2 )βˆ’(1/3^2 ))  =((11)/(54))+(1/3)(((49)/(36)))βˆ’(Ο€^2 /(18))=((71)/(108))βˆ’(Ο€^2 /(18))

∫01x2log(x)log(1βˆ’x)dx=Ξ½β€²(3,1)∫01xaβˆ’1(1βˆ’x)bβˆ’1dx=Ξ“(a)Ξ“(b)Ξ“(a+b)=Ξ½(a,b)∫01xaβˆ’1(1βˆ’x)bβˆ’1log(1βˆ’x)log(x)=Ξ½β€²(a,b)=βˆ‚βˆ‚a(Ξ“(a)Ξ“β€²(1)βˆ’Οˆ(a+1)Ξ“(1)Ξ“(a)Ξ“(a+1))=βˆ‚βˆ‚a(βˆ’Ξ³aβˆ’Οˆ(a+1)a)=Ξ³a2+ψ(1+a)a2βˆ’Οˆβ€²(a+1)a∣a=3=Ξ³9+ψ(4)9βˆ’Οˆβ€²(4)a=Ξ³βˆ’Ξ³+1+12+139βˆ’13(Ο€26βˆ’1βˆ’122βˆ’132)=1154+13(4936)βˆ’Ο€218=71108βˆ’Ο€218

Commented by mnjuly1970 last updated on 31/Mar/21

  thank you so much mr payan..

thankyousomuchmrpayan..

Commented by mnjuly1970 last updated on 31/Mar/21

 very nice..

verynice..

Terms of Service

Privacy Policy

Contact: info@tinkutara.com