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Question Number 51552 by maxmathsup by imad last updated on 28/Dec/18
calculate ∫_0 ^(+∞) ((arctan(1+x^2 ))/(x^2 +4))dx
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$
Commented by Abdo msup. last updated on 29/Dec/18
let I =∫_0 ^∞ ((arctan(1+x^2 ))/(x^2 +4))dx I =_(x=2t) ∫_0 ^∞ ((arctan(1+4t^2 ))/(4(1+t^2 )))2dt =(1/2) ∫_0 ^∞ ((arctan(1+4t^2 ))/(1+t^2 ))dt let consider f(x)=∫_0 ^∞ ((arctan(1+xt^2 ))/(1+t^2 ))dt with x>0 f^′ (x) =∫_0 ^∞ (t^2 /((1+x^2 t^4 )(1+t^2 )))dt ⇒2f(x)=∫_(−∞) ^(+∞) (t^2 /((x^2 t^4 +1)(t^2 +1)))dt let ϕ(z)=(z^2 /((x^2 z^4 +1)(z^2 +1))) ⇒ ϕ(z)= (z^2 /((xz^2 −i)(xz^2 +i)(z−i)(z+i))) =(z^2 /(x^2 (z^2 −(i/x))(z^2 +(i/x))(z−i)(z+i))) = (z^2 /(x^2 (z−(1/( (√x)))e^((iπ)/4) )(z +(1/( (√)x))e^((iπ)/4) )(z−(i/( (√x)))e^(−((iπ)/4)) )(z+(i/( (√x)))e^(−((iπ)/4)) )(z−i)(z+i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(1/( (√x)))e^((iπ)/4) )+Res(ϕ,−(1/( (√x)))e^(−((iπ)/4)) ) +Res(ϕ,i)} ...be continued....
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\: \\ $$$${I}\:=_{{x}=\mathrm{2}{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} \right)}{\mathrm{4}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\mathrm{2}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{let}\:{consider} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\Rightarrow\mathrm{2}{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} {t}^{\mathrm{4}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left({xz}^{\mathrm{2}} −{i}\right)\left({xz}^{\mathrm{2}} \:+{i}\right)\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} −\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}\:+\frac{\mathrm{1}}{\:\sqrt{}{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{{i}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{{i}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{i}\right)\left({z}+{i}\right)} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,{i}\right)\right\} \\ $$$$…{be}\:{continued}…. \\ $$

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