Question Number 39374 by maxmathsup by imad last updated on 05/Jul/18
$${calculate}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by math khazana by abdo last updated on 07/Jul/18
$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({tx}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$$${f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \:{x}^{\mathrm{4}} \right)}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{4}} \right)}{dx} \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {z}^{\mathrm{4}} \right)} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right){t}^{\mathrm{2}} \left({z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}^{\mathrm{2}} \:−\frac{{i}}{{t}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{t}}\right)} \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}\:−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\left({z}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\left({z}−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\left({z}+\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\:+{Res}\left(\varphi,−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\:\frac{−\mathrm{1}}{{t}^{\mathrm{2}} \left(\mathrm{2}{i}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\:=\frac{−\mathrm{1}}{\mathrm{2}{i}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${Res}\left(\varphi,\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\:=\:\:\frac{\frac{{i}}{{t}}}{{t}^{\mathrm{2}} \left(\mathrm{1}+\frac{{i}}{{t}}\right)\left(\mathrm{2}\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\left(\mathrm{2}\frac{{i}}{{t}}\right)} \\ $$$$=\:\:\frac{{i}}{\mathrm{4}{t}^{\mathrm{3}} \left(\:\frac{{t}+{i}}{{t}}\right)\:\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\:\frac{{i}}{{t}}}\:=\frac{{i}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \sqrt{{t}}}{\mathrm{4}{t}\left({t}+{i}\right){i}}\:\:=\frac{\sqrt{{t}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{t}\left({t}+{i}\right)} \\ $$$${Res}\left(\varphi,−\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\:=\:\frac{−\frac{{i}}{{t}}}{{t}^{\mathrm{2}} \left(−\frac{{i}}{{t}}\:+\mathrm{1}\right)\left(−\mathrm{2}\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\right)\left(\frac{−\mathrm{2}{i}}{{t}}\right)} \\ $$$$=\:\:\:\frac{−\mathrm{1}}{\mathrm{4}{t}^{\mathrm{3}} \left(\:\frac{−{i}+{t}}{{t}}\right)\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{t}}}\:\frac{\mathrm{1}}{{t}}}\:\:=\:−\frac{\sqrt{{t}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{t}\left({t}−{i}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\frac{\sqrt{{t}}}{\mathrm{4}{t}}\left(\:−\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{t}−{i}}\:+\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{{t}+{i}}\right)\right\} \\ $$$$=\:\frac{−\pi}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\:−\frac{{i}\pi\sqrt{{t}}}{\mathrm{2}{t}}\:\mathrm{2}{iIm}\left(\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{t}−{i}}\right)\:{but} \\ $$$$\:\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{t}−{i}}\:=\:\frac{\left({t}+{i}\right)\:\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{t}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{t}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:\Rightarrow\:{Im}\left(…\right) \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=−\frac{\pi}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\:\frac{\pi}{\:\sqrt{\mathrm{2}}\sqrt{{t}}}\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\mathrm{2}{f}^{'} \left({t}\right)\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\:−\frac{\pi}{\mathrm{2}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\:+\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({t}\right)\:\:=\:−\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:+{c} \\ $$
Commented by math khazana by abdo last updated on 07/Jul/18
$${but}\:\:\int_{\mathrm{0}} ^{{t}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:={arctant} \\ $$$$\int_{\mathrm{0}} ^{{t}} \:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{t}} \:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+\int_{\mathrm{0}} ^{{t}} \:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{arctant}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=−\frac{\pi}{\mathrm{2}}{arctan}\left({t}\right)\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{actant}\:+{c} \\ $$$${f}\left({t}\right)=\frac{\pi}{\mathrm{2}}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\right){arctan}\left({t}\right)\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$${c}={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\left(\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:−\frac{\pi}{\mathrm{2}}\right){arctan}\left({t}\right)\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:−\frac{\pi}{\mathrm{2}}\right)\:+\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}{\mathrm{16}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:+\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\left(\mathrm{2}\right)\:. \\ $$$$= \\ $$