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Question Number 137991 by EnterUsername last updated on 08/Apr/21

∫_0 ^(π/2) ln^3 (sinx)dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}^{\mathrm{3}} \left({sinx}\right){dx} \\ $$

Answered by Ar Brandon last updated on 08/Apr/21

f(α)=∫_0 ^(π/2) sin^(α−1) xdx=(1/2)β((α/2),(1/2))=(((√π)Γ((α/2)))/(2Γ(((α+1)/2)))) , f(1)=(π/2)★ ln(f(α))=ln(Γ((α/2)))−ln(Γ(((α+1)/2)))+ln(((√π)/2)) ((f ′(α))/(f(α)))=((Γ′((α/2)))/(2Γ((α/2))))−((Γ′(((α+1)/2)))/(2Γ(((α+1)/2))))=(1/2)ψ((α/2))−(1/2)ψ(((α+1)/2)) f ′(1)=(π/4)(ψ((1/2))−ψ(1))=(π/4)(−γ−2ln2+γ)=−((πln2)/2)★ f ′′(α)=((f ′(α))/2)(ψ((α/2))−ψ(((α+1)/2)))+((f(α))/4)(ψ′((α/2))−ψ′(((α+1)/2))) f ′′(1)=−((πln2)/4)(ψ((1/2))−ψ(1))+(π/8)(ψ′((1/2))−ψ′(1)) =−((πln2)/4)(−γ−2ln2+γ)+(π/8)(3ζ(2)−ζ(2))=((πln^2 2)/2)+(π^3 /(24))★ f ′′′(α)=((f ′′(α))/2)(ψ((α/2))−ψ(((α+1)/2)))+((f ′(α))/4)(ψ′((α/2))−ψ′(((α+1)/2))) +((f ′(α))/4)(ψ′((α/2))−ψ′(((α+1)/2)))+((f(α))/8)(ψ′′((α/2))−ψ′′(((α+1)/2))) f ′′′(1)=(((πln^2 2)/4)+(π^3 /(48)))(−2ln2)−((πln2)/8)((π^2 /2)−(π^2 /6)) −((πln2)/8)((π^2 /2)−(π^2 /6))+(π/(16))(−16×(7/8)ζ(3)+2ζ(3)) =−((πln^3 2)/2)−((π^3 ln2)/(24))−((π^3 ln2)/(16))+((π^3 ln2)/(48))−((π^3 ln2)/(16))+((π^3 ln2)/(48))−((3π)/4)ζ(3) =−((πln^3 2)/2)−((π^3 ln2)/8)−((3π)/4)ζ(3)★ ∫_0 ^(π/2) ln(sinx)dx=f ′(1), ∫_0 ^(π/2) ln^2 (sinx)dx=f ′′(1), ∫_0 ^(π/2) ln^3 (sinx)dx=f ′′′(1)

$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha−\mathrm{1}} \mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\alpha}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}\:,\:\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\bigstar \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}−\frac{\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\gamma−\mathrm{2ln2}+\gamma\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{2}}\bigstar \\ $$$$\mathrm{f}\:''\left(\alpha\right)=\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:''\left(\mathrm{1}\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)+\frac{\pi}{\mathrm{8}}\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi'\left(\mathrm{1}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\left(−\gamma−\mathrm{2ln2}+\gamma\right)+\frac{\pi}{\mathrm{8}}\left(\mathrm{3}\zeta\left(\mathrm{2}\right)−\zeta\left(\mathrm{2}\right)\right)=\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}}+\frac{\pi^{\mathrm{3}} }{\mathrm{24}}\bigstar \\ $$$$\mathrm{f}\:'''\left(\alpha\right)=\frac{\mathrm{f}\:''\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{8}}\left(\psi''\left(\frac{\alpha}{\mathrm{2}}\right)−\psi''\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:'''\left(\mathrm{1}\right)=\left(\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{4}}+\frac{\pi^{\mathrm{3}} }{\mathrm{48}}\right)\left(−\mathrm{2ln2}\right)−\frac{\pi\mathrm{ln2}}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\pi\mathrm{ln2}}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)+\frac{\pi}{\mathrm{16}}\left(−\mathrm{16}×\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln}^{\mathrm{3}} \mathrm{2}}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{24}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{16}}+\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{48}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{16}}+\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{48}}−\frac{\mathrm{3}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln}^{\mathrm{3}} \mathrm{2}}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{8}}−\frac{\mathrm{3}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right)\bigstar \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:'\left(\mathrm{1}\right),\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:''\left(\mathrm{1}\right),\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{3}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:'''\left(\mathrm{1}\right) \\ $$

Commented by Ar Brandon last updated on 08/Apr/21

ψ(α)=−γ+Σ_(n=0) ^∞ ((1/(n+1))−(1/(n+α))) ψ′(α)=Σ_(n=0) ^∞ (1/((n+α)^2 )) , ψ′′(α)=Σ_(n=0) ^∞ ((−2)/((n+α)^3 )) ⇒ψ′(1)=Σ_(n=0) ^∞ (1/((n+1)^2 ))=Σ_(n=1) ^∞ (1/n^2 )=ζ(2)=(π^2 /6) ⇒ψ′((1/2))=Σ_(n=0) ^∞ (1/((n+(1/2))^2 ))=4Σ_(n=0) ^∞ (1/((2n+1)^2 ))=4×(π^2 /8)=(π^2 /2) Σ_(n=0) ^∞ (1/((2n+1)^2 ))=1+(1/3^2 )+(1/5^2 )+... =(1+(1/2^2 )+(1/3^2 )+(1/4^2 )+...)−((1/2^2 )+(1/4^2 )+(1/6^2 )+...) =(1+(1/2^2 )+(1/3^2 )+(1/4^2 )+...)−(1/2^2 )(1+(1/2^2 )+(1/3^2 )+...) =ζ(2)−(1/4)ζ(2)=(3/4)ζ(2)=(π^2 /8) ψ′′(1)=Σ_(n=0) ^∞ ((−2)/((n+1)^3 ))=−2Σ_(n=1) ^∞ (1/n^3 )=−2ζ(3) ψ′′((1/2))=−2Σ_(n=0) ^∞ (1/((n+(1/2))^3 ))=−16Σ_(n=0) ^∞ (1/((2n+1)^3 )) =−16(1+(1/3^3 )+(1/5^3 )+...) =−16[(1+(1/2^3 )+(1/3^3 )+(1/4^3 )+(1/5^3 )+..)−(1/2^3 )(1+(1/2^3 )+(1/3^3 )+(1/4^3 )+...)] =−16×(1−(1/2^3 ))ζ(3)=−14ζ(3)

$$\psi\left(\alpha\right)=−\gamma+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\alpha}\right) \\ $$$$\psi'\left(\alpha\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\alpha\right)^{\mathrm{2}} }\:,\:\psi''\left(\alpha\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}}{\left(\mathrm{n}+\alpha\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\psi'\left(\mathrm{1}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{4}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}×\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+... \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+...\right)−\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+...\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+...\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...\right) \\ $$$$=\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\psi''\left(\mathrm{1}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }=−\mathrm{2}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }=−\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\psi''\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }=−\mathrm{16}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=−\mathrm{16}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+...\right) \\ $$$$=−\mathrm{16}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+..\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+...\right)\right] \\ $$$$=−\mathrm{16}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)\zeta\left(\mathrm{3}\right)=−\mathrm{14}\zeta\left(\mathrm{3}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 08/Apr/21

Peace in Chaos..

$${Peace}\:{in}\:{Chaos}.. \\ $$

Commented by Ar Brandon last updated on 08/Apr/21

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