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Question Number 39023 by maxmathsup by imad last updated on 01/Jul/18

let g(x)= ∫_(−∞) ^(+∞) ((arctan(x(1+t^2 )))/(1+t^2 ))dt with x>0 find a simple form of g(x) .

$${let}\:{g}\left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{arctan}\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:{with}\:{x}>\mathrm{0} \\ $$ $${find}\:{a}\:{simple}\:{form}\:{of}\:{g}\left({x}\right)\:. \\ $$

Commented bymath khazana by abdo last updated on 08/Jul/18

we have g^′ (x) = ∫_(−∞) ^(+∞) (1/(1+x^2 (1+t^2 )^2 ))dt = ∫_(−∞) ^(+∞) (dt/((x(1+t^2 )−i)(x(1+t^2 ) +i))) = ∫_(−∞) ^(+∞) (dt/((xt^2 +x−i)(xt^2 +x+i))) let ϕ(z) = (1/((xz^2 +x−i)(xz^2 +x+i))) ϕ(z) = (1/(x^2 (z^2 +1−(i/x))(z^2 +1 +(i/x)))) =(1/(x^2 ( z −i(√(1−(i/x))))(z +i(√(1−(i/x))))(z−i(√(1+(i/x))))(z+i(√(1+(i/x)))))) we have ∣1−(i/x)∣ =(√(1+(1/x^2 ))) =((√(x^2 +1))/x) 1−(i/x) =((√(x^2 +1))/x)( (x/(√(x^(2 ) +1))) −(i/x) (x/(√(x^2 +1)))) =r e^(iθ ) ⇒ r =((√(x^2 +1))/x) and cosθ = (x/(√(x^2 +1))) sin θ =−(1/(√(x^2 +1))) ⇒ tanθ = −(1/x) ⇒ θ =−arctan(x) ⇒1−(i/x) =r(x) e^(−i arctan((1/x))) (√(1−(i/x)))=(√(r(x)))e^(−(i/2)arctan((1/x))) and (√(1+(i/x)))= (√(r(x))) e^((i/2)arctan((1/x))) i(√(1−(i/x))) =(√(r(x))) e^(i(π/2)−(i/2)arctan((1/x))) =(√(r(x)))=(√(r(x))) e^(i( (π/2) −(1/2)((π/2) −arctanx))) =(√(r(x))) e^(i((π/4) +(1/2)arctan(x))) also i(√(1+(i/x)))=(√(r(x))) e^(i(π/2) +(i/2)arctan((1/x))) =(√(r(x))) e^(i((π/2) + (1/2)((π/2) −arctanx))) =(√(r(x))) e^(i( ((3π)/4) −(1/2) actan(x))) ....be continued...

$${we}\:{have}\:{g}^{'} \left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$ $$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dt}}{\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)−{i}\right)\left({x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{i}\right)} \\ $$ $$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\:\:\frac{{dt}}{\left({xt}^{\mathrm{2}} \:+{x}−{i}\right)\left({xt}^{\mathrm{2}} \:+{x}+{i}\right)} \\ $$ $${let}\:\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({xz}^{\mathrm{2}} \:+{x}−{i}\right)\left({xz}^{\mathrm{2}} \:+{x}+{i}\right)} \\ $$ $$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+\mathrm{1}−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:\:+\mathrm{1}\:+\frac{{i}}{{x}}\right)} \\ $$ $$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\:{z}\:−{i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\right)\left({z}\:+{i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\right)\left({z}−{i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\right)\left({z}+{i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}\right)} \\ $$ $${we}\:{have}\:\mid\mathrm{1}−\frac{{i}}{{x}}\mid\:=\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}} \\ $$ $$\mathrm{1}−\frac{{i}}{{x}}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\left(\:\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}\:} \:+\mathrm{1}}}\:−\frac{{i}}{{x}}\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\right) \\ $$ $$={r}\:{e}^{{i}\theta\:} \:\:\Rightarrow\:{r}\:=\frac{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}{{x}}\:\:{and}\:\:{cos}\theta\:=\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$ $${sin}\:\theta\:=−\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\:\:\Rightarrow\:{tan}\theta\:=\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$ $$\theta\:=−{arctan}\left({x}\right)\:\Rightarrow\mathrm{1}−\frac{{i}}{{x}}\:={r}\left({x}\right)\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$ $$\sqrt{\mathrm{1}−\frac{{i}}{{x}}}=\sqrt{{r}\left({x}\right)}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:{and} \\ $$ $$\sqrt{\mathrm{1}+\frac{{i}}{{x}}}=\:\sqrt{{r}\left({x}\right)}\:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$ $${i}\sqrt{\mathrm{1}−\frac{{i}}{{x}}}\:=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\frac{\pi}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$ $$=\sqrt{{r}\left({x}\right)}=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\left(\:\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctanx}\right)\right)} \\ $$ $$=\sqrt{{r}\left({x}\right)}\:\:{e}^{{i}\left(\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left({x}\right)\right)} \:\:{also} \\ $$ $${i}\sqrt{\mathrm{1}+\frac{{i}}{{x}}}=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\frac{\pi}{\mathrm{2}}\:+\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$ $$=\sqrt{{r}\left({x}\right)}\:\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}\:−{arctanx}\right)\right)} \\ $$ $$=\sqrt{{r}\left({x}\right)}\:{e}^{{i}\left(\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:{actan}\left({x}\right)\right)} \:\:\:\:\:....{be}\:{continued}... \\ $$ $$ \\ $$

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