Question Number 137991 by EnterUsername last updated on 08/Apr/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}^{\mathrm{3}} \left({sinx}\right){dx} \\ $$
Answered by Ar Brandon last updated on 08/Apr/21
$$\mathrm{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha−\mathrm{1}} \mathrm{xdx}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\alpha}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}\:,\:\mathrm{f}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}\bigstar \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\alpha\right)\right)=\mathrm{ln}\left(\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)\right)−\mathrm{ln}\left(\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\mathrm{ln}\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{f}\left(\alpha\right)}=\frac{\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}−\frac{\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{f}\:'\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)=\frac{\pi}{\mathrm{4}}\left(−\gamma−\mathrm{2ln2}+\gamma\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{2}}\bigstar \\ $$$$\mathrm{f}\:''\left(\alpha\right)=\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:''\left(\mathrm{1}\right)=−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\left(\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}\right)\right)+\frac{\pi}{\mathrm{8}}\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi'\left(\mathrm{1}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\left(−\gamma−\mathrm{2ln2}+\gamma\right)+\frac{\pi}{\mathrm{8}}\left(\mathrm{3}\zeta\left(\mathrm{2}\right)−\zeta\left(\mathrm{2}\right)\right)=\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{2}}+\frac{\pi^{\mathrm{3}} }{\mathrm{24}}\bigstar \\ $$$$\mathrm{f}\:'''\left(\alpha\right)=\frac{\mathrm{f}\:''\left(\alpha\right)}{\mathrm{2}}\left(\psi\left(\frac{\alpha}{\mathrm{2}}\right)−\psi\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{f}\:'\left(\alpha\right)}{\mathrm{4}}\left(\psi'\left(\frac{\alpha}{\mathrm{2}}\right)−\psi'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right)+\frac{\mathrm{f}\left(\alpha\right)}{\mathrm{8}}\left(\psi''\left(\frac{\alpha}{\mathrm{2}}\right)−\psi''\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\mathrm{f}\:'''\left(\mathrm{1}\right)=\left(\frac{\pi\mathrm{ln}^{\mathrm{2}} \mathrm{2}}{\mathrm{4}}+\frac{\pi^{\mathrm{3}} }{\mathrm{48}}\right)\left(−\mathrm{2ln2}\right)−\frac{\pi\mathrm{ln2}}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\pi\mathrm{ln2}}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)+\frac{\pi}{\mathrm{16}}\left(−\mathrm{16}×\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\mathrm{2}\zeta\left(\mathrm{3}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln}^{\mathrm{3}} \mathrm{2}}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{24}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{16}}+\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{48}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{16}}+\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{48}}−\frac{\mathrm{3}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi\mathrm{ln}^{\mathrm{3}} \mathrm{2}}{\mathrm{2}}−\frac{\pi^{\mathrm{3}} \mathrm{ln2}}{\mathrm{8}}−\frac{\mathrm{3}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right)\bigstar \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:'\left(\mathrm{1}\right),\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:''\left(\mathrm{1}\right),\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}^{\mathrm{3}} \left(\mathrm{sinx}\right)\mathrm{dx}=\mathrm{f}\:'''\left(\mathrm{1}\right) \\ $$
Commented by Ar Brandon last updated on 08/Apr/21
$$\psi\left(\alpha\right)=−\gamma+\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{n}+\alpha}\right) \\ $$$$\psi'\left(\alpha\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\alpha\right)^{\mathrm{2}} }\:,\:\psi''\left(\alpha\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}}{\left(\mathrm{n}+\alpha\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\psi'\left(\mathrm{1}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\psi'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{4}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}×\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+… \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+…\right)−\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+…\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }+…\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…\right) \\ $$$$=\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\psi''\left(\mathrm{1}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{−\mathrm{2}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{3}} }=−\mathrm{2}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }=−\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\psi''\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }=−\mathrm{16}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=−\mathrm{16}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+…\right) \\ $$$$=−\mathrm{16}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+..\right)−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+…\right)\right] \\ $$$$=−\mathrm{16}×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\right)\zeta\left(\mathrm{3}\right)=−\mathrm{14}\zeta\left(\mathrm{3}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 08/Apr/21
$${Peace}\:{in}\:{Chaos}.. \\ $$
Commented by Ar Brandon last updated on 08/Apr/21
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