Question Number 137894 by EnterUsername last updated on 08/Apr/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Apr/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{a}} \left({x}\right)=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:}{\mathrm{2}\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}=\tau\left({a}\right) \\ $$$$\tau''\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}^{\mathrm{2}} \left({sinx}\right){dx} \\ $$$$\tau'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sinx}\right){dx}=−\frac{\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right) \\ $$$${log}\left(\tau\left({a}\right)\right)={log}\left(\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\right)+{log}\left(\sqrt{\pi}/\mathrm{2}\right)−{log}\left(\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right) \\ $$$$\frac{\tau'\left({a}\right)}{\tau\left({a}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Rightarrow\tau''\left({a}\right)=\frac{\tau'\left({a}\right)}{\mathrm{2}}\left(\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right)+\frac{\tau\left({a}\right)}{\mathrm{4}}\left(\psi'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi'\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\right) \\ $$$$\Rightarrow\tau''\left(\mathrm{0}\right)=\frac{\tau'\left(\mathrm{0}\right)}{\mathrm{2}}\left(−\gamma+\gamma−{log}\left(\mathrm{4}\right)\right)+\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$=\frac{−\pi}{\mathrm{2}}{log}\left(\mathrm{2}\right)\left(−{log}\left(\mathrm{4}\right)\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{24}}=\frac{\pi}{\mathrm{2}}{log}^{\mathrm{2}} \left(\mathrm{2}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$
Commented by Dwaipayan Shikari last updated on 08/Apr/21
$${May}\:{be}\:{some}\:{error}.\:{Trying}\:{to}\:{do}\:{this}\:{again} \\ $$
Commented by Dwaipayan Shikari last updated on 08/Apr/21
$${Fixed}! \\ $$
Commented by Dwaipayan Shikari last updated on 08/Apr/21
$${If}\:\:{f}\left(\alpha\right)=\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\Rightarrow{f}'\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\psi\left(\frac{\alpha}{\mathrm{2}}\right) \\ $$$${From}\:{the}\:{definition}\:{of}\:{Digamma}\:{function} \\ $$$$\frac{\Gamma'\left(\alpha\right)}{\Gamma\left(\alpha\right)}=\psi\left(\alpha\right)=−\gamma+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\alpha} \\ $$
Commented by EnterUsername last updated on 08/Apr/21
$$\frac{\pi^{\mathrm{3}} }{\mathrm{24}}+\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}} \\ $$
Commented by EnterUsername last updated on 08/Apr/21
$${Excuse}\:{me}\:{please},\:{if}\:{f}\left(\alpha\right)=\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\:{then}\:{what}\:{is}\:{f}\:''\left(\alpha\right)\:? \\ $$
Commented by EnterUsername last updated on 08/Apr/21
$${Thanks} \\ $$
Commented by EnterUsername last updated on 08/Apr/21
$${I}\:{meant}\:{f}\:''\left(\alpha\right)\:{instead}\:{of}\:{f}\:'\left(\alpha\right) \\ $$
Commented by Dwaipayan Shikari last updated on 08/Apr/21
$${Same}\:{method} \\ $$$${f}'\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)\:\Rightarrow{f}''\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\partial}{\partial{a}}\left(\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma'\left(\frac{{a}}{\mathrm{2}}\right)\psi\left(\frac{{a}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\psi'\left(\frac{{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{{a}}{\mathrm{2}}\right)\psi^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\psi'\left(\frac{{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}}{\mathrm{2}}\right) \\ $$
Commented by EnterUsername last updated on 08/Apr/21
$${Thanks}\:{for}\:{the}\:{inspiration} \\ $$
Answered by EnterUsername last updated on 08/Apr/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx} \\ $$$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha−\mathrm{1}} \mathrm{xdx}=\frac{\sqrt{\pi}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}−\frac{\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{ln}\left(\mathrm{f}\:'\left(\alpha\right)\right)−\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{ln2} \\ $$$$\frac{\mathrm{f}\:''\left(\alpha\right)}{\mathrm{f}\:'\left(\alpha\right)}−\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\:,\:\mathrm{f}\:'\left(\mathrm{1}\right)=\frac{\mathrm{f}\left(\mathrm{1}\right)}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\mathrm{2ln2}\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{2}} \\ $$$$\psi'\left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:,\:\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{2f}\:''\left(\mathrm{1}\right)}{−\pi\mathrm{ln2}}+\frac{\pi\mathrm{ln2}}{\pi}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}}{−\mathrm{2ln2}}\:\Rightarrow−\frac{\mathrm{2f}\:''\left(\mathrm{1}\right)}{\pi\mathrm{ln2}}+\mathrm{ln2}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12ln2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:''\left(\mathrm{1}\right)=\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}}+\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$