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Question Number 140194 by mnjuly1970 last updated on 05/May/21

$$$$$$pleaseintegrate::$$$$f\left(x\right)={\int }_0^1\left\{\frac{1}{z}log\left(\frac{z^2+2zcos\left(x\right)+1}{{(z+1)}^2}\right)\right\}dz$$$$$$

Answered by Dwaipayan Shikari last updated on 05/May/21

$${\int }_0^1\frac{log(z^2+2zcosx+1)}{z}-2\frac{log(z+1)}{z}dz$$$$={\int }_0^1\frac{log(1+{ze}^{ix})+log(1+{ze}^{-ix})}{z}-2\mathrm{\Sigma }\frac{{(-1)}^{n+1}}{n^2}$$$$=\mathrm{\Sigma }{\int }_0^1{(-1)}^{n+1}\frac{1}{n}(\frac{z^n{e}^{inx}}{z}+\frac{z^n{e}^{-inx}}{z})dz-\frac{{\pi }^2}{6}$$$$=2\mathrm{\Sigma }{(-1)}^{n+1}\frac{cos\left(nx\right)}{n^2}-\frac{{\pi }^2}{6}=2\xi -\frac{{\pi }^2}{6}=-\frac{x^2}{2}$$$${\int }_0^x\mathrm{\Sigma }{(-1)}^{n+1}\frac{sin\left(nx\right)}{n^2}dx=\frac{{\pi }^2}{12}-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}cos\left(nx\right)}{n^2}$$$$\Rightarrow {\int }_0^x\frac{x}{2}dx=\frac{{\pi }^2}{12}-\xi \Rightarrow \xi =\frac{{\pi }^2}{12}-\frac{x^2}{4}$$$$$$

Commented by mnjuly1970 last updated on 05/May/21

$$perfectandnicesolution$$$$thankyou...$$

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