Advanced-calculus-0-1-x-2-ln-x-ln-1-x-dx-

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

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

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

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

φ=∫_0 ^1 x^2 ln(x)ln(1−x)dx=(∂/∂α)∣_(α=0) −∫_0 ^1 x^(2+α) Σ_(n=1) ^∞ (x^n /n)dx =(∂/∂α)∣_(α=0) −Σ_(n=1) ^∞ (1/n)∫x^(2+α+n) dx=(∂/∂α)∣_(α=0) −Σ_(n=1) ^∞ (1/(n(3+α+n))) =Σ_(n=1) ^∞ (1/(n(n+3)^2 ))=Σ_(n=1) ^∞ ((1/(9n))−(1/(9(n+3)))−(1/(3(n+3)^2 ))) =(1/9)Σ_(n=1) ^∞ ((1/n)−(1/(n+3)))−(1/3)(Σ_(n=1) ^∞ (1/n^2 )−(1+(1/2^2 )+(1/3^2 ))) =(1/9)(1+(1/2)+(1/3))−(1/3)(ζ(2)−(1+(1/2^2 )+(1/3^2 ))) =((11)/(54))−(π^2 /(18))+((49)/(108))=((71)/(108))−(π^2 /(18))

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

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

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

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

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

Commented by mnjuly1970 last updated on 31/Mar/21

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

Commented by mnjuly1970 last updated on 31/Mar/21

$${grateful}… \\ $$

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

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

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

But , whenever you are calculating ∫_0 ^1 x^π (1−x)^e log(x)log(1−x) dx you need this ν′(e+1,π+1) ;−)

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

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

OK, bro. Thanks 😊😃

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

∫_0 ^1 x^2 log(x)log(1−x)dx=ν′(3,1) ∫_0 ^1 x^(a−1) (1−x)^(b−1) dx=((Γ(a)Γ(b))/(Γ(a+b)))=ν(a,b) ∫_0 ^1 x^(a−1) (1−x)^(b−1) log(1−x)log(x)=ν′(a,b) =(∂/∂a)(((Γ(a)Γ′(1)−ψ(a+1)Γ(1)Γ(a))/(Γ(a+1)))) =(∂/∂a)(((−γ)/a)−((ψ(a+1))/a))=(γ/a^2 )+((ψ(1+a))/a^2 )−((ψ′(a+1))/a) ∣_(a=3) =(γ/9)+((ψ(4))/9)−((ψ′(4))/a)=((γ−γ+1+(1/2)+(1/3))/9)−(1/3)((π^2 /6)−1−(1/2^2 )−(1/3^2 )) =((11)/(54))+(1/3)(((49)/(36)))−(π^2 /(18))=((71)/(108))−(π^2 /(18))

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

Commented by mnjuly1970 last updated on 31/Mar/21

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

Commented by mnjuly1970 last updated on 31/Mar/21

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

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