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Question Number 120775 by mnjuly1970 last updated on 02/Nov/20

$$...advancedcalculus...$$$$evaluate::$$$$\mathrm{\Phi }\stackrel{???}{=}{\int }_0^{1}ln\left(x\right){tan}^{-1}\left(x\right)dx$$$$...m.n.1970...$$

Answered by Dwaipayan Shikari last updated on 02/Nov/20

$$\int {tan}^{-1}xdx$$$$={xtan}^{-1}x-\int \frac{x}{1+x^2}$$$$={xtan}^{-1}x-log\left(\sqrt{1+x^2}\right)(logx{\int }_0^1{tan}^{-1}\left(x\right)dx=0)$$$$\mathrm{\Phi }=\left[log\left(x\right){\int }_0^1{tan}^{-1}x\right]-{\int }_0^1{tan}^{-1}x+\frac{1}{2}{\int }_0^1\frac{log(1+x^2)}{x}$$$$\mathrm{\Phi }=-[{xtan}^{-1}x-log\left(\sqrt{1+x^2}\right)]+\frac{1}{2}{\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{x}^{2n-1}.\frac{1}{n}$$$$\mathrm{\Phi }=-\frac{\pi }{4}+\frac{1}{2}log\left(2\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}{\int }_0^1{x}^{2n-1}.\frac{1}{n}dx$$$$\mathrm{\Phi }=-\frac{\pi }{4}+\frac{1}{2}log\left(2\right)+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{n+1}\frac{1}{2n^2}$$$$\mathrm{\Phi }=-\frac{\pi }{4}+\frac{1}{2}log\left(2\right)+\frac{{\pi }^2}{48}$$

Commented by mnjuly1970 last updated on 02/Nov/20

$$thankyousomuchmrdwaipayanexcellent$$

Commented by Dwaipayan Shikari last updated on 03/Nov/20

$$Withpleasure$$

Answered by mindispower last updated on 02/Nov/20

$$={\left[(xln\left(x\right)-x){tan}^{-}\left(x\right)\right]}_0^1-{\int }_0^1\frac{xln\left(x\right)-x}{1+x^2}dx$$$$=-{tan}^{-}\left(1\right)+{\int }_0^1\frac{x}{1+x^2}dx-{\int }_0^1\frac{x}{1+x^2}ln\left(x\right)dx$$$$-\frac{\pi }{4}+\frac{ln\left(2\right)}{2}-\sum _{k⩾0}{\int }_0^1{(-1)}^k{x}^{2k+1}ln\left(x\right)dx$$$$=-\frac{\pi }{4}+\frac{ln\left(2\right)}{2}-\sum _{k⩾0}{(-1)}^k(-\frac{1}{{(2k+2)}^2})$$$$=-\frac{\pi }{4}+\frac{ln\left(2\right)}{2}+\frac{1}{4}\sum _{k⩾0}\frac{{(-1)}^k}{{(k+1)}^2}$$$$=-\frac{\pi }{4}+\frac{ln\left(2\right)}{2}+\frac{1}{4}\left[\frac{{\pi }^2}{12}\right]$$$$=\frac{-12\pi +{\pi }^2+24ln\left(2\right)}{48}$$

Commented by mnjuly1970 last updated on 02/Nov/20

$$verynicesirmindspower$$$$grateful..$$

Commented by mindispower last updated on 03/Nov/20

$$thanxwithepleasurnicedaysir$$

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