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let-g-x-arctan-x-1-t-2-1-t-2-dt-with-x-gt-0-find-a-simple-form-of-g-x-




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 by math 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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