Question Number 38121 by maxmathsup by imad last updated on 22/Jun/18
$${let}\:{x}>\mathrm{0}\:{find}\:{F}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\: \\ $$
Commented by prof Abdo imad last updated on 23/Jun/18
$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} \right)}{dt} \\ $$$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)}\:.{poles}\:{of}\:\varphi? \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\: \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}^{\mathrm{2}} \:−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)} \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}\:−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i},\:\overset{−} {+}\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} ,\overset{−} {+}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)=\:\:\frac{−\mathrm{1}}{\mathrm{2}{ix}^{\mathrm{2}} \left(\mathrm{1}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 23/Jun/18
$${Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)=\:\:\frac{{i}}{{x}^{\mathrm{3}} \left(\frac{{i}}{{x}}+\mathrm{1}\right)\left(\frac{\mathrm{2}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2}{i}}{{x}}\right)} \\ $$$$=\:\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} \left(\frac{{i}}{{x}}+\mathrm{1}\right)\mathrm{4}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=\:\frac{\sqrt{{x}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({xi}\:+{x}^{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\:\:\frac{−{i}}{{x}^{\mathrm{3}} \left(−\frac{{i}}{{x}}+\mathrm{1}\right)\left(\frac{−\mathrm{2}{i}}{{x}}\right)\left(\frac{−\mathrm{2}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\:\:\frac{−\sqrt{{x}}}{{x}^{\mathrm{2}} \left(−\frac{{i}}{{x}}+\mathrm{1}\right)\mathrm{4}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=\:\frac{−\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}^{\mathrm{2}} −{ix}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\:\frac{\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{x}\left({x}+{i}\right)}\:−\frac{\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{x}\left({x}−{i}\right)}\right\} \\ $$$$=\frac{−\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}}\left\{\:\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{{x}+{i}}\:−\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}−{i}}\right\} \\ $$$$=\frac{−\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}}\left\{\:\frac{\left({x}−{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:−\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}\right\} \\ $$$$=−\frac{\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{2}{iIm}\left(\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$$=−\frac{\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{\pi\sqrt{{x}}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{Im}\left\{\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:{but} \\ $$$$\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} =\left({x}+{i}\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{i}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({x}+{i}\right)\left(\mathrm{1}+{i}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\:{x}\:+{xi}\:+{i}\:−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\:{x}−\mathrm{1}\:+{i}\left({x}+\mathrm{1}\right)\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=−\frac{\pi}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:+\frac{\pi\sqrt{{x}}}{\:\sqrt{\mathrm{2}}{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\left({x}+\mathrm{1}\right)={f}^{'} \left({x}\right) \\ $$$$\Rightarrow\:{f}\left({x}\right)=−\pi\:{arctan}\left({x}\right)\:+\frac{\pi}{\:\sqrt{\mathrm{2}}}\int_{.} ^{{x}} \:\frac{\left({t}+\mathrm{1}\right)\sqrt{{t}}}{{t}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:\:+{c} \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 23/Jun/18
$${changement}\:\sqrt{{t}}={u}\:{give} \\ $$$$\int_{.} ^{{x}} \:\:\:\frac{\left({t}+\mathrm{1}\right)\sqrt{{t}}}{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:=\:\int_{.} ^{\sqrt{{x}}} \:\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right){u}}{{u}^{\mathrm{2}} \left({u}^{\mathrm{4}} +\mathrm{1}\right)}\:\mathrm{2}{u}\:{du} \\ $$$$=\int_{.} ^{\sqrt{{x}}} \:\:\:\:\frac{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{\mathrm{1}+{u}^{\mathrm{4}} }\:{du}…{be}\:{cpntinued}… \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
$${let}\:{drcompose}\:{F}\left({u}\right)=\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{{u}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$${F}\left({u}\right)=\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\:{u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{u}+\mathrm{1}\right)} \\ $$$$\:=\frac{{au}\:+{b}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}+\mathrm{1}}={F}\left({u}\right) \\ $$$$\Rightarrow{c}=−{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${F}\left({u}\right)=\:\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{2}=\:\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{1}\:\Rightarrow \\ $$$${F}\left({u}\right)=\:\frac{{au}\:+\mathrm{1}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}+\mathrm{1}}\:+\frac{−{au}\:+\mathrm{1}}{{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)=\mathrm{2}\:=\frac{{a}+\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{−{a}+\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$$$=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}\:+\mathrm{2}−\sqrt{\mathrm{2}}\:−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}\:+\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{2}\sqrt{\mathrm{2}}{a}\:\:+\mathrm{4}}{\mathrm{2}}\:=−\sqrt{\mathrm{2}}{a}\:+\mathrm{2}=\mathrm{2}\:\Rightarrow{a}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({u}\right)=\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{u}\:+\mathrm{1}} \\ $$$$\int\:{F}\left({u}\right)\:{du}\:=\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$+\:\int\:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:−\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\int\:\:\:\:\frac{{du}}{\left({u}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:+\int\:\:\:\:\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$={I}\:+{J} \\ $$$${I}\:=_{{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{t}} \:\:\:\:\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\frac{{dt}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}\:{arctan}\left({u}\sqrt{\mathrm{2}}\:+\mathrm{1}\right) \\ $$$${J}=_{{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{t}} \:\:\:\sqrt{\mathrm{2}}\:{arctan}\left({u}\sqrt{\mathrm{2}}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\int_{.} ^{\sqrt{{x}}} \:{F}\left({u}\right){du}\:=\:\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right)+\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
$$\Rightarrow{F}\left({x}\right)=−\pi\:{arctan}\left({x}\right)\:+\pi\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right) \\ $$$$+\pi\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:−\mathrm{1}\right)\:+\lambda \\ $$$$\lambda=\:{lim}_{{x}\rightarrow\mathrm{0}} {F}\left({x}\right)=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)=\pi\left\{\:{arctan}\left(\sqrt{\mathrm{2}{x}}+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right)\:−{arctanx}\right\} \\ $$