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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 byprof 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 byprof 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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