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Question Number 33259 by prof Abdo imad last updated on 14/Apr/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2x\right)}{a^2+x^2}dxwitha\ne 0$$

Commented by prof Abdo imad last updated on 27/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(2x\right)}{a^2+x^2}dx$$$$ifa>0ch.x=atgive$$$$I={\int }_0^{\mathrm{\infty }}\frac{arctan\left(2at\right)}{a^2(1+t^2)}adt=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{arctan\left(2at\right)}{1+t^2}dt$$$$\Rightarrow aI={\int }_0^{\mathrm{\infty }}\frac{arctan\left(2at\right)}{1+t^2}dt=f\left(a\right)wehave$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{2t}{(1+4a^2{t}^2)(1+t^2)}dtletdecompose$$$$F\left(t\right)=\frac{2t}{(1+4a^2{t}^2)(1+t^2)}=\frac{\alpha t+b}{t^2+1}+\frac{ct+d}{4a^2{t}^2+1}$$$$F(-t)=-F\left(t\right)\Rightarrow \frac{-\alpha t+b}{t^2+1}+\frac{-ct+d}{4a^2{t}^2+1}$$$$=\frac{-\alpha t-b}{t^2+1}+\frac{-ct-d}{4a^2{t}^2+1}\Rightarrow b=d=0\Rightarrow$$$$F\left(t\right)=\frac{\alpha t}{t^2+1}+\frac{ct}{4a^2{t}^2+1}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=\alpha +\frac{c}{4a^2}\Rightarrow 4a^2\alpha +c=0\Rightarrow$$$$c=-4a^2\alpha \Rightarrow F\left(t\right)=\frac{\alpha t}{t^2+1}-4a^2\frac{\alpha t}{4a^2{t}^2+1}$$$$F\left(1\right)=\frac{2}{(1+4a^2)2}=\frac{1}{4a^2+1}=\frac{\alpha }{2}-\frac{4a^2\alpha }{4a^2+1}\Rightarrow$$$$1=\frac{1}{2}(4a^2+1)\alpha -4a^2\alpha =(2a^2+\frac{1}{2}-4a^2)\alpha$$$$=(\frac{1}{2}-2a^2)\alpha =\frac{1-4a^2}{2}\alpha \Rightarrow \alpha =\frac{2}{1-4a^2}$$$$F\left(t\right)=\frac{2}{1-4a^2}\frac{t}{t^2+1}-4a^2\frac{2}{1-4a^2}\frac{t}{4a^2{t}^2+1}$$$$F\left(t\right)=\frac{2}{1-4a^2}\frac{t}{t^2+1}-\frac{8a^2}{1-4a^2}\frac{t}{4a^2{t}^2+1}$$$$f^{{}^{\prime }}\left(a\right)=\frac{1}{1-4a^2}{\int }_0^{\mathrm{\infty }}\frac{2tdt}{t^2+1}-\frac{8a^2}{1-4a^2}{\int }_0^{\mathrm{\infty }}\frac{tdt}{4a^2{t}^2+1}$$$$but{\int }_0^{\mathrm{\infty }}\frac{tdt}{4a^2{t}^2+1}=\frac{1}{8a^2}{\int }_0^{\mathrm{\infty }}\frac{8a^{2}t}{4a^2{t}^2+1}$$$$f^{{}^{\prime }}\left(a\right)=\frac{1}{1-4a^2}{\left[ln\left(\frac{1+t^2}{4a^2{t}^2+1}\right)\right]}_0^{+\mathrm{\infty }}=\frac{1}{1-4a^2}ln\left(\frac{1}{4a^2}\right)$$$$=\frac{-ln\left(4a^2\right)}{1-4a^2}\Rightarrow f\left(a\right)={\int }_0^a\frac{-ln\left(4x^2\right)}{1-4x^2}dx+\lambda$$$$but\lambda =f\left(0\right)=0\Rightarrow f\left(a\right)=-{\int }_0^a\frac{2ln\left(2x\right)}{1-4x^2}dx$$$$-f\left(a\right){=}_{2x=t}2{\int }_0^{2a}\frac{ln\left(t\right)}{1-t^2}\frac{dt}{2}={\int }_0^{2a}\frac{ln\left(t\right)}{1-t^2}dt$$$$if0<2a<1\iff 0<a<\frac{1}{2}$$$${\int }_0^{2a}\frac{ln\left(t\right)}{1-t^2}dt={\int }_0^{2a}\left(\sum _{n=0}^{\mathrm{\infty }}{t}^{2n}\right)ln\left(t\right)dt$$$$=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{2a}{t}^{2n}ln\left(t\right)dt=\sum _{n=0}^{\mathrm{\infty }}{A}_n$$$$A_n={\int }_0^{2a}{t}^{2n}ln\left(t\right)dtbecalculatedbyrecurrence$$$$....becontinued....$$$$$$

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