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Question Number 154080 by iloveisrael last updated on 14/Sep/21

$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}\ln {}^2\left(\frac{1+\sin t}{1-\sin t}\right)dt$$

Answered by mindispower last updated on 14/Sep/21

$$={\int }_0^{\frac{\pi }{2}}4{ln}^2\left(\frac{1+tg\left(\frac{x}{2}\right)}{1-tg\left(\frac{x}{2}\right)}\right)dx$$$$y=tg\left(\frac{x}{2}\right)\Rightarrow dx=\frac{2dy}{1+y^2}$$$$=8{\int }_0^1\frac{{ln}^2\left(\frac{1-y}{1+y}\right)}{1+y^2}dy$$$$y=\frac{1-x}{1+x}\Rightarrow dy=\frac{-2dx}{{(1+x)}^2}$$$$\mathrm{\Omega }=8{\int }_0^1\frac{{ln}^2\left(x\right)}{1+{\left(\frac{1-x}{1+x}\right)}^2}.\frac{2dx}{{(1+x)}^2}=8{\int }_0^1\frac{{ln}^2\left(x\right)}{1+x^2}dx$$$$=8{\int }_0^1\sum _{n⩾0}{(-x^2)}^n{ln}^2\left(x\right)dx$$$$=8\sum _{n⩾0}{(-1)}^n{\int }_0^1{x}^{2n}{ln}^2\left(x\right)dx$$$${\int }_0^1{x}^{2n}{ln}^2\left(x\right)dx={\int }_0^{\mathrm{\infty }}{t}^2{e}^{-(2n+1)t}dt$$$$\frac{1}{{(2n+1)}^3}{\int }_0^{\mathrm{\infty }}{t}^2{e}^{-t}=\frac{2}{{(1+2n)}^3}$$$$\mathrm{\Omega }=16\sum _{n⩾0}\frac{{(-1)}^n}{{(1+2n)}^3}$$$$=16\sum _{n⩾0}(\frac{1}{{(1+4n)}^3}-\frac{1}{{(4n+3)}^3})$$$$=\frac{16}{64}\sum _{n⩾0}(\frac{1}{{(n+\frac{1}{4})}^3}-\frac{1}{{(n+\frac{3}{4})}^3})$$$$=\frac{16}{64}({\mathrm{\Psi }}^3\left(\frac{1}{4}\right)-{\mathrm{\Psi }}^3\left(\frac{3}{4}\right))$$$$$$$$$$

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