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Question Number 137894 by EnterUsername last updated on 08/Apr/21

∫_0 ^(π/2) ln^2 (sinx)dx

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

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

∫_0 ^(π/2) sin^a (x)=((Γ(((a+1)/2))Γ((1/2)) )/(2Γ((a/2)+1)))=τ(a) τ′′(0)=∫_0 ^(π/2) log^2 (sinx)dx τ′(0)=∫_0 ^(π/2) log(sinx)dx=−(π/2)log(2) log(τ(a))=log(Γ(((a+1)/2)))+log((√π)/2)−log(Γ((a/2)+1)) ((τ′(a))/(τ(a)))=(1/2)ψ(((a+1)/2))−ψ((a/2)+1)⇒τ′′(a)=((τ′(a))/2)(ψ(((a+1)/2))−ψ((a/2)+1))+((τ(a))/4)(ψ′(((a+1)/2))−ψ′((a/2)+1)) ⇒τ′′(0)=((τ′(0))/2)(−γ+γ−log(4))+(π/8)((π^2 /2)−(π^2 /6)) =((−π)/2)log(2)(−log(4))+(π^3 /(24))=(π/2)log^2 (2)+(π^3 /(24))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({x}\right)=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:}{\mathrm{2}\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}=\tau\left({a}\right) \\ $$$$\tau''\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}^{\mathrm{2}} \left({sinx}\right){dx} \\ $$$$\tau'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx}=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$${log}\left(\tau\left({a}\right)\right)={log}\left(\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\right)+{log}\left(\sqrt{\pi}/\mathrm{2}\right)−{log}\left(\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right) \\ $$$$\frac{\tau'\left({a}\right)}{\tau\left({a}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Rightarrow\tau''\left({a}\right)=\frac{\tau'\left({a}\right)}{\mathrm{2}}\left(\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right)+\frac{\tau\left({a}\right)}{\mathrm{4}}\left(\psi'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi'\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right) \\ $$$$\Rightarrow\tau''\left(\mathrm{0}\right)=\frac{\tau'\left(\mathrm{0}\right)}{\mathrm{2}}\left(−\gamma+\gamma−{log}\left(\mathrm{4}\right)\right)+\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$=\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)\left(−{log}\left(\mathrm{4}\right)\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{24}}=\frac{\pi}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$

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

May be some error. Trying to do this again

$${May}\:{be}\:{some}\:{error}.\:{Trying}\:{to}\:{do}\:{this}\:{again} \\ $$

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

Fixed!

$${Fixed}! \\ $$

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

If f(α)=Γ((α/2))⇒f′(α)=(1/2)Γ′((α/2))=(1/2)Γ((α/2))ψ((α/2)) From the definition of Digamma function ((Γ′(α))/(Γ(α)))=ψ(α)=−γ+Σ_(n=0) ^∞ (1/(n+1))−(1/(n+α))

$${If}\:\:{f}\left(\alpha\right)=\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\Rightarrow{f}'\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\psi\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$${From}\:{the}\:{definition}\:{of}\:{Digamma}\:{function} \\ $$$$\frac{\Gamma'\left(\alpha\right)}{\Gamma\left(\alpha\right)}=\psi\left(\alpha\right)=−\gamma+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\alpha} \\ $$

Commented by EnterUsername last updated on 08/Apr/21

(π^3 /(24))+((πln^2 2)/2)

$$\frac{\pi^{\mathrm{3}} }{\mathrm{24}}+\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}} \\ $$

Commented by EnterUsername last updated on 08/Apr/21

Excuse me please, if f(α)=Γ((α/2)) then what is f ′′(α) ?

$${Excuse}\:{me}\:{please},\:{if}\:{f}\left(\alpha\right)=\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\:{then}\:{what}\:{is}\:{f}\:''\left(\alpha\right)\:? \\ $$

Commented by EnterUsername last updated on 08/Apr/21

Thanks

$${Thanks} \\ $$

Commented by EnterUsername last updated on 08/Apr/21

I meant f ′′(α) instead of f ′(α)

$${I}\:{meant}\:{f}\:''\left(\alpha\right)\:{instead}\:{of}\:{f}\:'\left(\alpha\right) \\ $$

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

Same method f′(a)=(1/2)Γ((a/2))ψ((a/2)) ⇒f′′(a)=(1/2).(∂/∂a)(Γ((a/2))ψ((a/2))) =(1/4)Γ′((a/2))ψ((a/2))+(1/4)ψ′((a/2))Γ((a/2)) =(1/4)Γ((a/2))ψ^2 ((a/2))+(1/4)ψ′((a/2))Γ((a/2))

$${Same}\:{method} \\ $$$${f}'\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)\:\Rightarrow{f}''\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\partial}{\partial{a}}\left(\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma'\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\psi'\left(\frac{{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\psi'\left(\frac{{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}}{\mathrm{2}}\right) \\ $$

Commented by EnterUsername last updated on 08/Apr/21

Thanks for the inspiration

$${Thanks}\:{for}\:{the}\:{inspiration} \\ $$

Answered by EnterUsername last updated on 08/Apr/21

∫_0 ^(π/2) ln^2 (sinx)dx f(α)=∫_0 ^(π/2) sin^(α−1) xdx=(((√π)Γ((α/2)))/(2Γ(((α+1)/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)) ln(f ′(α))−ln(f(α))=ln(ψ((α/2))−ψ(((α+1)/2)))−ln2 ((f ′′(α))/(f ′(α)))−((f ′(α))/(f(α)))=(1/2)∙((ψ′((α/2))−ψ′(((α+1)/2)))/(ψ((α/2))−ψ(((α+1)/2)))) f(1)=(π/2) , f ′(1)=((f(1))/2)(ψ((1/2))−ψ(1))=(π/4)(−2ln2)=−((πln2)/2) ψ′(1)=ζ(2)=(π^2 /6) , ψ′((1/2))=(π^2 /2) ⇒((2f ′′(1))/(−πln2))+((πln2)/π)=(1/2)∙(((π^2 /2)−(π^2 /6))/(−2ln2)) ⇒−((2f ′′(1))/(πln2))+ln2=−(π^2 /(12ln2)) ⇒∫_0 ^(π/2) ln^2 (sinx)dx=f ′′(1)=((πln^2 2)/2)+(π^3 /(24))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx} \\ $$$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha−\mathrm{1}} \mathrm{xdx}=\frac{\sqrt{\pi}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}−\frac{\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{ln}\left(\mathrm{f}\:'\left(\alpha\right)\right)−\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln2} \\ $$$$\frac{\mathrm{f}\:''\left(\alpha\right)}{\mathrm{f}\:'\left(\alpha\right)}−\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\:,\:\mathrm{f}\:'\left(\mathrm{1}\right)=\frac{\mathrm{f}\left(\mathrm{1}\right)}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\mathrm{2ln2}\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{2}} \\ $$$$\psi'\left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:,\:\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{2f}\:''\left(\mathrm{1}\right)}{−\pi\mathrm{ln2}}+\frac{\pi\mathrm{ln2}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}}{−\mathrm{2ln2}}\:\Rightarrow−\frac{\mathrm{2f}\:''\left(\mathrm{1}\right)}{\pi\mathrm{ln2}}+\mathrm{ln2}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12ln2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:''\left(\mathrm{1}\right)=\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}}+\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$

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