Question Number 39023 by maxmathsup by imad last updated on 01/Jul/18
$${let}\:{g}\left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:{with}\:{x}>\mathrm{0} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{g}\left({x}\right)\:. \\ $$
Commented by math khazana by abdo last updated on 08/Jul/18
$${we}\:{have}\:{g}^{'} \left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dt}}{\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−{i}\right)\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{i}\right)} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\frac{{dt}}{\left({xt}^{\mathrm{2}} \:+{x}−{i}\right)\left({xt}^{\mathrm{2}} \:+{x}+{i}\right)} \\ $$$${let}\:\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({xz}^{\mathrm{2}} \:+{x}−{i}\right)\left({xz}^{\mathrm{2}} \:+{x}+{i}\right)} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+\mathrm{1}−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:\:+\mathrm{1}\:+\frac{{i}}{{x}}\right)} \\ $$$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\:{z}\:−{i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\right)\left({z}\:+{i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\right)\left({z}−{i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\right)\left({z}+{i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\right)} \\ $$$${we}\:{have}\:\mid\mathrm{1}−\frac{{i}}{{x}}\mid\:=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}} \\ $$$$\mathrm{1}−\frac{{i}}{{x}}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\left(\:\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}\:} \:+\mathrm{1}}}\:−\frac{{i}}{{x}}\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\right) \\ $$$$={r}\:{e}^{{i}\theta\:} \:\:\Rightarrow\:{r}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\:\:{and}\:\:{cos}\theta\:=\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$${sin}\:\theta\:=−\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\:\:\Rightarrow\:{tan}\theta\:=\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$$\theta\:=−{arctan}\left({x}\right)\:\Rightarrow\mathrm{1}−\frac{{i}}{{x}}\:={r}\left({x}\right)\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$\sqrt{\mathrm{1}−\frac{{i}}{{x}}}=\sqrt{{r}\left({x}\right)}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:{and} \\ $$$$\sqrt{\mathrm{1}+\frac{{i}}{{x}}}=\:\sqrt{{r}\left({x}\right)}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\:=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\frac{\pi}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\sqrt{{r}\left({x}\right)}=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\left(\:\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctanx}\right)\right)} \\ $$$$=\sqrt{{r}\left({x}\right)}\:\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left({x}\right)\right)} \:\:{also} \\ $$$${i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\frac{\pi}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\sqrt{{r}\left({x}\right)}\:\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctanx}\right)\right)} \\ $$$$=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\left(\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{actan}\left({x}\right)\right)} \:\:\:\:\:….{be}\:{continued}… \\ $$$$ \\ $$