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Question Number 197099 by Erico last updated on 07/Sep/23

$${\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\ln \left(\cos t\right)}{\sin t}\mathrm{d}t=???$$

Answered by witcher3 last updated on 07/Sep/23

$$={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\mathrm{t}\right)\ln \left(\cos \left(\mathrm{t}\right)\right)}{1-{\cos }^2\left(\mathrm{t}\right)}\mathrm{dt},\cos \left(\mathrm{t}\right)=\mathrm{y}$$$$={\int }_0^1\frac{\ln \left(\mathrm{y}\right)}{1-{\mathrm{y}}^2}\mathrm{dt}=\sum _{\mathrm{n}⩾0}{\int }_0^1{\mathrm{y}}^{2\mathrm{n}}\ln \left(\mathrm{t}\right)\mathrm{dt}$$$$=-\sum _{\mathrm{n}⩾0}\frac{1}{{(2\mathrm{n}+1)}^2}-(\zeta \left(2\right)-\frac{1}{4}\zeta \left(2\right))=-\frac{{\pi }^2}{8}$$

Answered by Mathspace last updated on 09/Sep/23

$$tan\left(\frac{t}{2}\right)=x\Rightarrow$$$$I={\int }_0^1\frac{ln\left(\frac{1-x^2}{1+x^2}\right)}{\frac{2x}{1+x^2}}\frac{2dx}{1+x^2}$$$$={\int }_0^1\frac{ln(1-x^2)-ln(1+x^2)}{x}dx$$$${ln}^{{}^{\prime }}(1-u)=\frac{-1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n$$$$\Rightarrow ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}$$$$\Rightarrow ln(1-x^2)=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n}$$$${ln}^{{}^{\prime }}(1+u)=\frac{1}{1+u}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{u}^n$$$$\Rightarrow ln(1+u)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{u^{n+1}}{n+1}$$$$=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{u^n}{n}and$$$$ln(1+x^2)=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n-1}\frac{x^{2n}}{n}$$$$\Rightarrow \frac{ln(1-x^2)-ln(1+x^2)}{x}$$$$=-\frac{1}{x}\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2n}}{n}+\frac{1}{x}\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n\frac{x^{2n}}{n}$$$$=\sum _{n=1}^{\mathrm{\infty }}({(-1)}^n-1)\frac{x^{2n-1}}{n}$$$$=-2\sum _{n=1}^{\mathrm{\infty }}\frac{x^{2(2n+1)-1}}{2n+1}$$$$=-2\sum _{n=1}^{\mathrm{\infty }}\frac{x^{4n+1}}{2n+1}and$$$$I=-2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n+1)}{\int }_0^1{x}^{4n+1}dx$$$$=-2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2n+1)(4n+2)}$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}$$$$\mathrm{\Sigma }\frac{1}{n^2}=\frac{1}{4}\mathrm{\Sigma }\frac{1}{n^2}+\mathrm{\Sigma }\frac{1}{{(2n+1)}^2}$$$$\Rightarrow \frac{3}{4}\mathrm{\Sigma }\frac{1}{n^2}=\mathrm{\Sigma }\frac{1}{{(2n+1)}^2}$$$$\Rightarrow \frac{3}{4}.\frac{{\pi }^2}{6}=\mathrm{\Sigma }\frac{1}{{(2n+1)}^2}$$$$\Rightarrow \frac{{\pi }^2}{8}=\mathrm{\Sigma }\frac{1}{{(2n+1)}^2}\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}\frac{ln\left(cosx\right)}{sinx}=-\frac{{\pi }^2}{8}$$

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