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Question Number 43058 by maxmathsup by imad last updated on 06/Sep/18
calculate ∫_0 ^∞ ((x sin(((πx)/2)))/({1+(x+1)^2 }{1+(x−1)^2 }))dx
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{x}\:{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\left\{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right\}\left\{\mathrm{1}+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 09/Sep/18
2I = ∫_(−∞) ^(+∞) ((xsin(((πx)/2)))/({1+(x+1)^2 }{1+(x−1)^2 }))dx=Im (∫_(−∞) ^(+∞) ((x e^((iπx)/2) )/({1+(x+1)^2 }{1+(x−1)^2 }))dx) let ϕ(z) = ((z e^((iπz)/2) )/({1+(z+1)^2 }{1+(z−1)^2 })) we have ϕ(z) = ((z e^((iπz)/2) )/({1+z^2 +2z+1}{1+z^2 −2z +1})) = ((z e^((iπz)/2) )/({z^2 +2z+2}{z^2 −2z +2})) poles of ϕ ? roots of z^2 +2z +2→ Δ^′ = 1−2 =−1 ⇒ z_1 =−1+i and z_2 =−1−i roots of z^2 −2z +2 → Δ^′ =1−2=−1 ⇒z_3 =1+i and z_4 =1−i ⇒ ϕ(z) = ((z e^((iπz)/2) )/((z−z_1 )(z−z_2 )(z−z_3 )(z−z_4 ))) = ((z e^((iπz)/2) )/((z−z_1 )(z−z_2 )(z+z_2 )(z+z_1 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,z_1 )+Res(ϕ,z_3 )} Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) = ((z_1 e^((iπz_1 )/2) )/((z_1 −z_2 )(z_1 +z_2 )(2z_1 ))) = (e^((iπz_1 )/2) /(2(2i)(−2))) = (e^((iπz_1 )/2) /(−8i)) Res(ϕ, z_3 ) = Res(ϕ, −z_2 ) =lim_(z→−z_2 ) (z+z_2 )ϕ(z) = ((−z_2 e^((−iπz_2 )/2) )/((−z_2 −z_1 )(−2z_2 )(z_1 −z_2 ))) = (e^(−((iπz_2 )/2)) /(−2(z_1 +z_2 )(z_1 −z_2 ))) = (e^(−((iπz_2 )/2)) /((−2)(−2)(2i))) = (e^(−((iπz_2 )/2)) /(8i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ −(e^((iπz_1 )/2) /(8i)) + (e^(−((iπz_2 )/2)) /(8i))}=−(π/4){ e^((iπz_1 )/2) −e^(−((iπ z_1 ^− )/2)) } =−(π/4) {2i Im(e^((iπz_1 )/2) )} =−((iπ)/2) Im( e^((iπz_1 )/2) ) but e^((iπz_1 )/2) =e^((iπ(−1+i))/2) = e^((−iπ−π)/2) = e^(−(π/2)) (−i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2) (−e^(−(π/2)) ) = ((iπ)/2) e^(−(π/2)) ⇒ 2I = (π/2) e^(−(π/2)) ⇒ I = (π/4) e^(−(π/2)) .
$$\:\mathrm{2}{I}\:\:=\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{xsin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\left\{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right\}\left\{\mathrm{1}+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}}{dx}={Im}\:\left(\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{e}^{\frac{{i}\pi{x}}{\mathrm{2}}} }{\left\{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right\}\left\{\mathrm{1}+\left({x}−\mathrm{1}\right)^{\mathrm{2}} \right\}}{dx}\right) \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{\frac{{i}\pi{z}}{\mathrm{2}}} }{\left\{\mathrm{1}+\left({z}+\mathrm{1}\right)^{\mathrm{2}} \right\}\left\{\mathrm{1}+\left({z}−\mathrm{1}\right)^{\mathrm{2}} \right\}}\:\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{\frac{{i}\pi{z}}{\mathrm{2}}} }{\left\{\mathrm{1}+{z}^{\mathrm{2}} \:+\mathrm{2}{z}+\mathrm{1}\right\}\left\{\mathrm{1}+{z}^{\mathrm{2}} −\mathrm{2}{z}\:+\mathrm{1}\right\}}\:=\:\frac{{z}\:{e}^{\frac{{i}\pi{z}}{\mathrm{2}}} }{\left\{{z}^{\mathrm{2}} \:+\mathrm{2}{z}+\mathrm{2}\right\}\left\{{z}^{\mathrm{2}} \:−\mathrm{2}{z}\:+\mathrm{2}\right\}} \\ $$$${poles}\:{of}\:\varphi\:? \\ $$$${roots}\:{of}\:\:{z}^{\mathrm{2}} \:+\mathrm{2}{z}\:+\mathrm{2}\rightarrow\:\Delta^{'} \:=\:\mathrm{1}−\mathrm{2}\:=−\mathrm{1}\:\Rightarrow\:{z}_{\mathrm{1}} =−\mathrm{1}+{i}\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{1}−{i} \\ $$$${roots}\:{of}\:{z}^{\mathrm{2}} −\mathrm{2}{z}\:+\mathrm{2}\:\rightarrow\:\Delta^{'} \:=\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\Rightarrow{z}_{\mathrm{3}} =\mathrm{1}+{i}\:\:{and}\:{z}_{\mathrm{4}} =\mathrm{1}−{i}\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{\frac{{i}\pi{z}}{\mathrm{2}}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{3}} \right)\left({z}−{z}_{\mathrm{4}} \right)}\:=\:\frac{{z}\:{e}^{\frac{{i}\pi{z}}{\mathrm{2}}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}+{z}_{\mathrm{2}} \right)\left({z}+{z}_{\mathrm{1}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,{z}_{\mathrm{3}} \right)\right\} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right)\varphi\left({z}\right)\:=\:\frac{{z}_{\mathrm{1}} \:{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} }{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)\left(\mathrm{2}{z}_{\mathrm{1}} \right)} \\ $$$$=\:\frac{{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{2}{i}\right)\left(−\mathrm{2}\right)}\:=\:\frac{{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} }{−\mathrm{8}{i}} \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{3}} \right)\:=\:{Res}\left(\varphi,\:−{z}_{\mathrm{2}} \right)\:={lim}_{{z}\rightarrow−{z}_{\mathrm{2}} } \left({z}+{z}_{\mathrm{2}} \right)\varphi\left({z}\right) \\ $$$$=\:\frac{−{z}_{\mathrm{2}} \:{e}^{\frac{−{i}\pi{z}_{\mathrm{2}} }{\mathrm{2}}} }{\left(−\boldsymbol{{z}}_{\mathrm{2}} −\boldsymbol{{z}}_{\mathrm{1}} \right)\left(−\mathrm{2}\boldsymbol{{z}}_{\mathrm{2}} \right)\left(\boldsymbol{{z}}_{\mathrm{1}} −\boldsymbol{{z}}_{\mathrm{2}} \right)}\:=\:\frac{\boldsymbol{{e}}^{−\frac{\boldsymbol{{i}}\pi{z}_{\mathrm{2}} }{\mathrm{2}}} }{−\mathrm{2}\left({z}_{\mathrm{1}} \:+{z}_{\mathrm{2}} \right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{{e}^{−\frac{{i}\pi{z}_{\mathrm{2}} }{\mathrm{2}}} }{\left(−\mathrm{2}\right)\left(−\mathrm{2}\right)\left(\mathrm{2}\boldsymbol{{i}}\right)}\:=\:\frac{\boldsymbol{{e}}^{−\frac{\boldsymbol{{i}}\pi{z}_{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{8}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:−\frac{{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} }{\mathrm{8}{i}}\:+\:\frac{{e}^{−\frac{{i}\pi{z}_{\mathrm{2}} }{\mathrm{2}}} }{\mathrm{8}{i}}\right\}=−\frac{\pi}{\mathrm{4}}\left\{\:\:{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} \:−{e}^{−\frac{{i}\pi\:\overset{−} {{z}}_{\mathrm{1}} }{\mathrm{2}}} \right\} \\ $$$$=−\frac{\pi}{\mathrm{4}}\:\left\{\mathrm{2}{i}\:{Im}\left({e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} \right)\right\}\:=−\frac{{i}\pi}{\mathrm{2}}\:{Im}\left(\:{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} \right)\:\:{but} \\ $$$$\:{e}^{\frac{{i}\pi{z}_{\mathrm{1}} }{\mathrm{2}}} \:\:\:={e}^{\frac{{i}\pi\left(−\mathrm{1}+{i}\right)}{\mathrm{2}}} \:\:=\:{e}^{\frac{−{i}\pi−\pi}{\mathrm{2}}} \:=\:{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\left(−\boldsymbol{{i}}\right)\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=−\frac{{i}\pi}{\mathrm{2}}\:\left(−{e}^{−\frac{\pi}{\mathrm{2}}} \right) \\ $$$$=\:\frac{{i}\pi}{\mathrm{2}}\:{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:\:\Rightarrow\:\mathrm{2}{I}\:=\:\frac{\pi}{\mathrm{2}}\:{e}^{−\frac{\pi}{\mathrm{2}}} \:\Rightarrow\:{I}\:=\:\frac{\pi}{\mathrm{4}}\:{e}^{−\frac{\pi}{\mathrm{2}}} \:\:. \\ $$$$ \\ $$

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