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Question Number 141218 by mathmax by abdo last updated on 16/May/21

$$\mathrm{calculate}\mathrm{I}={\int }_0^{\frac{\pi }{2}}{}^4\sqrt{\mathrm{tanx}}\log \left(\mathrm{tanx}\right)\mathrm{dx}\mathrm{and}$$$$\mathrm{J}={\int }_0^{\frac{\pi }{2}}\frac{\log \left(\mathrm{tanx}\right)}{\left({}^3\sqrt{\mathrm{tanx}}\right)}\mathrm{dx}$$

Answered by Ar Brandon last updated on 16/May/21

$$\mathrm{f}\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\mathrm{xdx}=\frac{1}{2}\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)=\frac{\pi }{2\sin \left(\frac{1-\alpha }{2}\pi \right)}$$$$\mathrm{f}\left(\frac{1}{4}\right)=\frac{\mathrm{\Gamma }\left(\frac{5}{8}\right)\mathrm{\Gamma }\left(\frac{3}{8}\right)}{2}=\frac{\pi }{2\sin \left(\frac{3\pi }{8}\right)}★$$$$\ln \left(\mathrm{f}\left(\alpha \right)\right)=\ln \left(\frac{\pi }{2}\right)-\mathrm{lnsin}\left(\frac{1-\alpha }{2}\pi \right)$$$$\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{\mathrm{f}\left(\alpha \right)}=\frac{\pi }{2}\cdot \tan \left(\frac{1-\alpha }{2}\pi \right)\Rightarrow \mathrm{f}{}^{\prime }\left(\alpha \right)=\mathrm{f}\left(\alpha \right)\{\frac{\pi }{2}\cdot \tan \left(\frac{1-\alpha }{2}\pi \right)\}$$$$\mathrm{f}{}^{\prime }\left(\frac{1}{4}\right)=\mathrm{f}\left(\frac{1}{4}\right)\{\frac{\pi }{2}\cdot \tan \left(\frac{3\pi }{8}\right)\}$$$$\mathrm{I}=\frac{{\pi }^2}{4}\tan \left(\frac{3\pi }{8}\right)\mathrm{cosec}\left(\frac{3\pi }{8}\right)$$

Answered by Ar Brandon last updated on 16/May/21

$$\mathrm{f}\left(\alpha \right)={\int }_0^{\frac{\pi }{2}}{\tan }^{\alpha }\mathrm{xdx}=\frac{1}{2}\mathrm{\Gamma }\left(\frac{\alpha +1}{2}\right)\mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)=\frac{\pi }{2\sin \left(\frac{1+\alpha }{2}\pi \right)}$$$$\mathrm{f}(-\frac{1}{3})=\frac{\pi }{2\sin \left(\frac{\pi }{3}\right)}★$$$$\ln \left(\mathrm{f}\left(\alpha \right)\right)=\ln \left(\frac{\pi }{2}\right)-\mathrm{lnsin}\left(\frac{1+\alpha }{2}\pi \right)$$$$\frac{\mathrm{f}{}^{\prime }\left(\alpha \right)}{\mathrm{f}\left(\alpha \right)}=\frac{\pi }{2}\cdot \tan \left(\frac{1+\alpha }{2}\pi \right)\Rightarrow \mathrm{f}{}^{\prime }\left(\alpha \right)=\mathrm{f}\left(\alpha \right)\{\frac{\pi }{2}\cdot \tan \left(\frac{1+\alpha }{2}\pi \right)\}$$$$\mathrm{f}{}^{\prime }(-\frac{1}{3})=\mathrm{f}(-\frac{1}{3})\{\frac{\pi }{2}\cdot \tan \left(\frac{\pi }{3}\right)\}$$$$\mathrm{J}=\frac{{\pi }^2}{4}\tan \left(\frac{\pi }{3}\right)\mathrm{cosec}\left(\frac{\pi }{3}\right)$$

Commented by Ar Brandon last updated on 16/May/21

$$\mathrm{f}{}^{\prime }\left(\alpha \right)=\frac{{\pi }^2}{4}\tan \left(\frac{1-\alpha }{2}\pi \right)\mathrm{cosec}\left(\frac{1-\alpha }{2}\pi \right)$$

Answered by mathmax by abdo last updated on 16/May/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{\mathrm{a}}\mathrm{dx}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{\mathrm{aln}\left(\mathrm{tanx}\right)}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{\mathrm{a}}\log \left(\mathrm{tanx}\right)\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{4}\right)={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{\frac{1}{4}}\log \left(\mathrm{tanx}\right)\mathrm{dx}=\mathrm{I}$$$$\mathrm{changement}\mathrm{tanx}=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{a}}}{1+{\mathrm{t}}^2}\mathrm{dt}{=}_{\mathrm{t}=\sqrt{\mathrm{z}}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{\mathrm{a}}{2}}}{1+\mathrm{z}}\frac{\mathrm{dz}}{2\sqrt{\mathrm{z}}}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{\mathrm{a}}{2}-\frac{1}{2}}}{1+\mathrm{z}}\mathrm{dz}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\frac{\mathrm{a}+1}{2}-1}}{1+\mathrm{z}}\mathrm{dz}=\frac{\pi }{2\sin \left(\pi \left(\frac{\mathrm{a}+1)}{2}\right)\right)}$$$$=\frac{\pi }{2\cos \left(\frac{\pi \mathrm{a}}{2}\right)}\mathrm{cv}\mathrm{of}\mathrm{f}\left(\mathrm{a}\right)\mathrm{we}\mathrm{must}\mathrm{have}0<\frac{\mathrm{a}+1}{2}<1\Rightarrow -1<\mathrm{a}<1$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{\pi }{2}\times \frac{-\frac{\pi }{2}\sin \left(\frac{\pi \mathrm{a}}{2}\right)}{{\cos }^2\left(\frac{\pi \mathrm{a}}{2}\right)}=\frac{{\pi }^2}{4}.\frac{\sin \left(\frac{\pi \mathrm{a}}{2}\right)}{{\cos }^2\left(\frac{\pi \mathrm{a}}{2}\right)}$$$$\mathrm{I}={\mathrm{f}}^{{}^{\prime }}\left(\frac{1}{4}\right)=\frac{{\pi }^2}{4}.\frac{\sin \left(\frac{\pi }{8}\right)}{{\cos }^2\left(\frac{\pi }{8}\right)}=\frac{{\pi }^2}{4}.\frac{\frac{\sqrt{2-\sqrt{2}}}{2}}{\frac{2+\sqrt{2}}{4}}=\frac{{\pi }^2}{2}\times \frac{\sqrt{2-\sqrt{2}}}{2+\sqrt{2}}$$$$\mathrm{J}={\int }_0^{\frac{\pi }{2}}{\left(\mathrm{tanx}\right)}^{-\frac{1}{3}}\log \left(\mathrm{tanx}\right)\mathrm{dx}={\mathrm{f}}^{{}^{\prime }}(-\frac{1}{3})$$$$=-\frac{{\pi }^2}{4}.\frac{\sin \left(\frac{\pi }{6}\right)}{{\cos }^2\left(\frac{\pi }{6}\right)}=-\frac{{\pi }^2}{4}\times \frac{\frac{1}{2}}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=-\frac{{\pi }^2}{8}\times \frac{4}{3}$$$$=-\frac{{\pi }^2}{6}$$

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