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

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

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\mathscr{A}{dvanced}\:\:...\:\:{calculus}...... \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} {ln}\left({x}\right){ln}\left(\mathrm{1}βˆ’{x}\right){dx}=??? \\ $$$$ \\ $$

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))

$$\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}βˆ’\mathrm{x}\right)\mathrm{dx}=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{2}+\alpha} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\mathrm{dx} \\ $$$$\:\:\:=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} βˆ’\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\int\mathrm{x}^{\mathrm{2}+\alpha+\mathrm{n}} \mathrm{dx}=\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{0}} βˆ’\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{3}+\alpha+\mathrm{n}\right)} \\ $$$$\:\:\:=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} }=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{9n}}βˆ’\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{n}+\mathrm{3}\right)}βˆ’\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{9}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}}βˆ’\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\right)βˆ’\frac{\mathrm{1}}{\mathrm{3}}\left(\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }βˆ’\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)βˆ’\frac{\mathrm{1}}{\mathrm{3}}\left(\zeta\left(\mathrm{2}\right)βˆ’\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)\right) \\ $$$$\:\:\:=\frac{\mathrm{11}}{\mathrm{54}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{18}}+\frac{\mathrm{49}}{\mathrm{108}}=\frac{\mathrm{71}}{\mathrm{108}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$

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

:)

$$\left.:\right) \\ $$

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

Even shorter :)

$$\left.{Even}\:{shorter}\::\right) \\ $$

Commented by mnjuly1970 last updated on 31/Mar/21

thanks alot mr brandon...

$${thanks}\:{alot}\:{mr}\:{brandon}... \\ $$

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 !

$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome},\:\mathrm{Sir}\:! \\ $$

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’)

$${But}\:,\:{whenever}\:{you}\:{are}\:{calculating}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\pi} \left(\mathrm{1}βˆ’{x}\right)^{{e}} {log}\left({x}\right){log}\left(\mathrm{1}βˆ’{x}\right)\:{dx} \\ $$$$\left.{you}\:{need}\:{this}\:\nu'\left({e}+\mathrm{1},\pi+\mathrm{1}\right)\:;βˆ’\right) \\ $$

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))

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {log}\left({x}\right){log}\left(\mathrm{1}βˆ’{x}\right){dx}=\nu'\left(\mathrm{3},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}βˆ’\mathrm{1}} \left(\mathrm{1}βˆ’{x}\right)^{{b}βˆ’\mathrm{1}} {dx}=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}=\nu\left({a},{b}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}βˆ’\mathrm{1}} \left(\mathrm{1}βˆ’{x}\right)^{{b}βˆ’\mathrm{1}} {log}\left(\mathrm{1}βˆ’{x}\right){log}\left({x}\right)=\nu'\left({a},{b}\right) \\ $$$$=\frac{\partial}{\partial{a}}\left(\frac{\Gamma\left({a}\right)\Gamma'\left(\mathrm{1}\right)βˆ’\psi\left({a}+\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)\Gamma\left({a}\right)}{\Gamma\left({a}+\mathrm{1}\right)}\right) \\ $$$$=\frac{\partial}{\partial{a}}\left(\frac{βˆ’\gamma}{{a}}βˆ’\frac{\psi\left({a}+\mathrm{1}\right)}{{a}}\right)=\frac{\gamma}{{a}^{\mathrm{2}} }+\frac{\psi\left(\mathrm{1}+{a}\right)}{{a}^{\mathrm{2}} }βˆ’\frac{\psi'\left({a}+\mathrm{1}\right)}{{a}}\:\mid_{{a}=\mathrm{3}} \\ $$$$=\frac{\gamma}{\mathrm{9}}+\frac{\psi\left(\mathrm{4}\right)}{\mathrm{9}}βˆ’\frac{\psi'\left(\mathrm{4}\right)}{{a}}=\frac{\gammaβˆ’\gamma+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{9}}βˆ’\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}βˆ’\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }βˆ’\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{11}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{49}}{\mathrm{36}}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{18}}=\frac{\mathrm{71}}{\mathrm{108}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 31/Mar/21

thank you so much mr payan..

$$\:\:{thank}\:{you}\:{so}\:{much}\:{mr}\:{payan}.. \\ $$

Commented by mnjuly1970 last updated on 31/Mar/21

very nice..

$$\:{very}\:{nice}.. \\ $$

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