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Question Number 97627 by mathmax by abdo last updated on 08/Jun/20

$$\mathrm{give}{\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}$$

Answered by mathmax by abdo last updated on 09/Jun/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{arctanx}\right)=\frac{1}{1+{\mathrm{x}}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}$$$$\Rightarrow \arctan \left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}+\mathrm{c}(\mathrm{c}=0)\Rightarrow$$$$\mathrm{A}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}{\mathrm{w}}_{\mathrm{n}}$$$${\mathrm{w}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\tan \theta \mathrm{give}$$$${\mathrm{w}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}\frac{{\tan }^{\mathrm{n}}\theta }{{(1+{\tan }^2\theta )}^2}(1+{\tan }^2\theta )\mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}{\cos }^2\theta {\tan }^{\mathrm{n}}\theta \mathrm{d}\theta$$$$\Rightarrow \mathrm{A}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{w}}_{\mathrm{n}}}{2\mathrm{n}+1}$$

Answered by mathmax by abdo last updated on 09/Jun/20

$$\mathrm{another}\mathrm{way}\mathrm{we}\mathrm{have}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^1\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}+{\int }_1^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}(\to \mathrm{x}=\frac{1}{\mathrm{t}})$$$$={\int }_0^1\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}+{\int }_0^1\frac{\frac{\pi }{2}-\mathrm{arctant}}{{(1+\frac{1}{{\mathrm{t}}^2})}^2}\left(\frac{\mathrm{dt}}{{\mathrm{t}}^2}\right)$$$$={\int }_0^1\frac{\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}+{\int }_0^1\frac{(\frac{\pi }{2}-\mathrm{arctant}){\mathrm{t}}^2}{{(1+{\mathrm{t}}^2)}^2}\mathrm{dt}$$$$={\int }_0^1\frac{\arctan \left(\mathrm{x}\right)+\frac{\pi }{2}{\mathrm{x}}^2-{\mathrm{x}}^2\mathrm{arctanx}}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^1\frac{(1-{\mathrm{x}}^2)\mathrm{arctanx}}{{(1+{\mathrm{x}}^2)}^2}+\frac{\pi }{2}{\int }_0^1\frac{{\mathrm{x}}^2\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}$$$$\mathrm{we}\mathrm{have}\mathrm{for}\mid \mathrm{u}\mid <1\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow -\frac{1}{{(1+\mathrm{u})}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}-1}$$$$\Rightarrow \frac{1}{{(1+\mathrm{u})}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-1)}^{\mathrm{n}-1}{\mathrm{u}}^{\mathrm{n}-1}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow$$$$\frac{1}{{(1+{\mathrm{x}}^2)}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\Rightarrow$$$${\int }_0^1\frac{{\mathrm{x}}^2\mathrm{dx}}{{(1+{\mathrm{x}}^2)}^2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}+2}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{(\mathrm{n}+1){(-1)}^{\mathrm{n}}}{2\mathrm{n}+3}$$$${\int }_0^1\frac{(1-{\mathrm{x}}^2)\arctan \left(\mathrm{x}\right)}{{(1+{\mathrm{x}}^2)}^2}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){(-1)}^{\mathrm{n}}{\int }_0^1(1-{\mathrm{x}}^2){\mathrm{x}}^{2\mathrm{n}}\arctan \left(\mathrm{x}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1){(-1)}^{\mathrm{n}}{\mathrm{A}}_{\mathrm{n}}\mathrm{with}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1({\mathrm{x}}^{2\mathrm{n}}-{\mathrm{x}}^{2\mathrm{n}+2})\mathrm{arctanx}\mathrm{dx}$$$${\mathrm{A}}_{\mathrm{n}}={\left[(\frac{1}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}-\frac{1}{2\mathrm{n}+3}{\mathrm{x}}^{2\mathrm{n}+3})\mathrm{arctanx}\right]}_0^1-{\int }_0^1(\frac{1}{2\mathrm{n}+1}{\mathrm{x}}^{2\mathrm{n}+1}-\frac{1}{2\mathrm{n}+3}{\mathrm{x}}^{2\mathrm{n}+3})\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\frac{\pi }{4}(\frac{1}{2\mathrm{n}+1}-\frac{1}{2\mathrm{n}+3})-\frac{1}{2\mathrm{n}+1}{\int }_0^1\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{1+{\mathrm{x}}^2}\mathrm{dx}+\frac{1}{2\mathrm{n}+3}{\int }_0^1\frac{{\mathrm{x}}^{2\mathrm{n}+3}}{1+{\mathrm{x}}^2}\mathrm{dx}...\mathrm{be}\mathrm{continued}...$$$$$$

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