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Question Number 146775 by Willson last updated on 15/Jul/21

$$\mathbf{Prove}\mathbf{that}$$$${\underset{0}{\int }}^{\mathit{\pi }}\mathit{t}\mathit{l}\mathit{n}\left(\mathit{s}\mathit{i}\mathit{n}\mathit{t}\right)\mathit{d}\mathit{t}=-\frac{{\mathit{\pi }}^2}{2}\mathit{l}\mathit{n}\left(2\right)$$

Answered by ArielVyny last updated on 15/Jul/21

$${\int }_0^{\pi }t(-ln2-\sum _{n⩾1}\frac{cos\left(2nt\right)}{n})dt$$$$-ln\left(2\right){\int }_0^{\pi }tdt-{\int }_0^{\pi }\left(t\sum _{n⩾1}\frac{cos\left(2nt\right)}{n}\right)dt$$$$-\frac{{\pi }^2}{2}ln2accordingthat{\int }_0^{\pi }t\sum _{n⩾1}\frac{cos\left(2nt\right)}{n}dt=0$$$${\int }_0^{\pi }tln\left(sint\right)dt=-\frac{{\pi }^{2}ln2}{2}$$$$$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$\mathrm{\Omega }={\int }_0^{\pi }t\ln \left(\sin t\right)dt\left(1\right)$$$$\mathrm{Let}u=\pi -t$$$$\mathrm{\Omega }={\int }_0^{\pi }(\pi -u)\ln \left(\sin u\right)du\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}:\mathrm{\Omega }=\frac{\pi }{2}{\int }_0^{\pi }\ln \left(\sin t\right)dt\left(3\right)$$$$\mathrm{\Omega }=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\sin t\right)dt+\frac{\pi }{2}{\int }_{\frac{\pi }{2}}^{\pi }\ln \left(\sin t\right)dt$$$$\mathrm{Let}u=t-\frac{\pi }{2}$$$$\mathrm{\Omega }=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\sin t\right)dt+\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\ln \left(\cos u\right)du\left(4\right)$$$$$$$$\mathrm{Let}t=2u$$$$\left(3\right):\mathrm{\Omega }=\pi {\int }_0^{\frac{\pi }{2}}\ln \left(\sin \left(2u\right)\right)du$$$$\mathrm{\Omega }=\pi {\int }_0^{\frac{\pi }{2}}\ln 2+\ln \left(\sin u\right)+\ln \left(\cos u\right)du\left(5\right)$$$$$$$$\left(5\right)-2\times \left(4\right)w:$$$$-\mathrm{\Omega }=\pi {\int }_0^{\frac{\pi }{2}}\ln 2du=\frac{{\pi }^2}{2}\ln \left(2\right)$$$$\Rightarrow \mathrm{\Omega }=-\frac{{\pi }^2}{2}\ln \left(2\right)$$

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

$$\mathrm{{\rm Y}}={\int }_0^{\pi }\mathrm{t}\log \left(\mathrm{sint}\right)\mathrm{dt}\Rightarrow \mathrm{{\rm Y}}={\int }_0^{\frac{\pi }{2}}\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}+{\int }_{\frac{\pi }{2}}^{\pi }\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}(\to \mathrm{t}=\frac{\pi }{2}+\mathrm{z})$$$$={\int }_0^{\frac{\pi }{2}}\mathrm{tlog}\left(\mathrm{sint}\right)\mathrm{dt}[+{\int }_0^{\frac{\pi }{2}}(\frac{\pi }{2}+\mathrm{z})\log \left(\mathrm{cost}\right)\mathrm{dt}$$$$=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\log \left(\mathrm{cost}\right)\mathrm{dt}+{\int }_0^{\frac{\pi }{2}}\mathrm{t}(\log \left(\mathrm{sint}\right)+\log \left(\mathrm{cost}\right))\mathrm{dt}$$$$=-\frac{{\pi }^2}{4}\log 2+{\int }_0^{\frac{\pi }{2}}\mathrm{tlog}\left(\frac{\sin \left(2\mathrm{t}\right)}{2}\right)\mathrm{dt}$$$$=-\frac{{\pi }^2}{4}\log 2+{\int }_0^{\frac{\pi }{2}}\mathrm{tlog}(\sin \left(2\mathrm{t}\right)\mathrm{dt}(\to 2\mathrm{t}=\mathrm{u})-\log 2{\int }_0^{\frac{\pi }{2}}\mathrm{tdt}$$$$=-\frac{{\pi }^2}{4}\log 2+{\int }_0^{\pi }\frac{\mathrm{u}}{2}\log \left(\mathrm{sinu}\right)\frac{\mathrm{du}}{2}-\log 2\times \frac{{\pi }^2}{8}$$$$=-\frac{3{\pi }^2}{8}\log 2+\frac{\mathrm{{\rm Y}}}{4}\Rightarrow (1-\frac{1}{4})\mathrm{{\rm Y}}=-\frac{3{\pi }^2}{8}\log 2\Rightarrow$$$$\frac{3}{4}\mathrm{{\rm Y}}=-\frac{3{\pi }^2}{8}\log 2\Rightarrow \mathrm{{\rm Y}}=-\frac{{\pi }^2}{2}\log 2$$$$$$

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