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Question Number 33341 by prof Abdo imad last updated on 14/Apr/18
find the value of ∫_0 ^∞ (dx/((1+x^2 )(a^2 +x^2 ))) 2) find the value of A(θ) = ∫_0 ^∞ (dx/((1+x^2 )( x^2 +1 −sin^2 θ))) 0<θ<(π/2) .
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:{x}^{\mathrm{2}} \:+\mathrm{1}\:−{sin}^{\mathrm{2}} \theta\right)} \\ $$$$\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:. \\ $$
Commented by prof Abdo imad last updated on 18/Apr/18
1) let put f(a) = ∫_0 ^∞ (dx/((1+x^2 )(a^2 +x^2 ))) 2f(a) = ∫_(−∞) ^(+∞) (dx/((1+x^2 )(a^2 +x^2 ))) let vonsider the complex function ϕ(z)= (1/((z^2 +1)(z^(2 ) +a^2 ))) the poles of ϕ is i,−i,ia ,−ia case 1 a>0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,ia)) but ϕ(z) = (1/((z−i)(z+i)(z−ia)(z+ia))) ⇒ Res(ϕ,i) = (1/((2i)(a^2 −1))) Res(ϕ,ia) = (1/((2ia)(1−a^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( (1/((−2i)(1−a^2 ))) +(1/((2ia)(1−a^2 )))) = ((2π)/(1−a^2 ))( (1/(2a)) −(1/2)) = (π/(1−a^2 ))( (1/a) −1) = (π/(1−a^2 )) ((1−a)/a) = (π/(a(1+a))) ⇒ I = (π/(2a(1+a))) case2 a <0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,−ia)) Res(ϕ,−ia) = (1/((−2ia)(1−a^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( ((−1)/(2i(1−a^2 ))) −(1/(2ia( 1−a^2 )))) = ((−π)/(1−a^2 ))( 1+ (1/a)) = ((−π(1+a))/(a(1−a^2 ))) = ((−π)/(a(1−a))) = (π/(a(a−1))) ⇒ I = (π/(2a(a−1))) and we must study the special case a =+^− 1 .
$$\left.\mathrm{1}\right)\:{let}\:{put}\:\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)} \\ $$$$\mathrm{2}{f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)}\:\:{let}\:{vonsider} \\ $$$${the}\:{complex}\:{function}\:\varphi\left({z}\right)=\:\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{2}\:} \:+{a}^{\mathrm{2}} \right)} \\ $$$${the}\:{poles}\:{of}\:\:\varphi\:{is}\:{i},−{i},{ia}\:,−{ia} \\ $$$${case}\:\mathrm{1}\:\:\:{a}>\mathrm{0}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{ia}\right)\right)\:{but} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{ia}\right)\left({z}+{ia}\right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)\left({a}^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$$${Res}\left(\varphi,{ia}\right)\:=\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{i}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:+\frac{\mathrm{1}}{\left(\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$$$=\:\frac{\mathrm{2}\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\:\frac{\mathrm{1}}{\mathrm{2}{a}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\frac{\mathrm{1}}{{a}}\:−\mathrm{1}\right) \\ $$$$=\:\frac{\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\:\frac{\mathrm{1}−{a}}{{a}}\:=\:\frac{\pi}{{a}\left(\mathrm{1}+{a}\right)}\:\:\:\Rightarrow\:{I}\:=\:\:\frac{\pi}{\mathrm{2}{a}\left(\mathrm{1}+{a}\right)} \\ $$$${case}\mathrm{2}\:\:{a}\:<\mathrm{0}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,−{ia}\right)\right) \\ $$$${Res}\left(\varphi,−{ia}\right)\:=\:\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{ia}\right)\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:−\frac{\mathrm{1}}{\mathrm{2}{ia}\left(\:\mathrm{1}−{a}^{\mathrm{2}} \right)}\right) \\ $$$$=\:\frac{−\pi}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\:\:\mathrm{1}+\:\frac{\mathrm{1}}{{a}}\right)\:=\:\frac{−\pi\left(\mathrm{1}+{a}\right)}{{a}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:=\:\:\frac{−\pi}{{a}\left(\mathrm{1}−{a}\right)} \\ $$$$=\:\:\frac{\pi}{{a}\left({a}−\mathrm{1}\right)}\:\Rightarrow\:{I}\:\:=\:\:\frac{\pi}{\mathrm{2}{a}\left({a}−\mathrm{1}\right)}\:\:\:{and}\:{we}\:{must} \\ $$$${study}\:{the}\:{special}\:{case}\:\:{a}\:=\overset{−} {+}\:\mathrm{1}\:. \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 18/Apr/18
2) we have A(θ) = ∫_0 ^∞ (dx/((1+x^2 )( x^2 +cos^2 θ))) =f (cosθ) = (π/(2 cosθ(1+cosθ))) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{A}\left(\theta\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:{x}^{\mathrm{2}} \:+{cos}^{\mathrm{2}} \theta\right)} \\ $$$$={f}\:\left({cos}\theta\right)\:=\:\:\frac{\pi}{\mathrm{2}\:{cos}\theta\left(\mathrm{1}+{cos}\theta\right)}\:. \\ $$

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