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Question Number 33835 by prof Abdo imad last updated on 25/Apr/18

find the value of ∫_(−∞) ^(+∞) ((cos(πx))/((x^2 +1+i)^2 )) dx

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{1}+{i}\right)^{\mathrm{2}} }\:{dx} \\ $$

Commented by prof Abdo imad last updated on 29/Apr/18

let consider the complex function ϕ(z)= (e^(iπz) /((z^2 +1+i)^2 )) we have z^2 +1+i =z^2 −(−1−i) −1−i =(√2) (((−1)/(√2)) −(i/(√2))) =(√2)(cos(π+(π/4)) +i sin(π+(π/4))) =(√2) e^(i((5π)/4)) ⇒ z^2 −(−1−i) =z^2 −(√2) e^(i((5π)/4)) =(z −^4 ((√2) e^(i((5π)/8)) )^2 so the poles are 2^(1/4) e^(i((5π)/8)) =z_0 and −2^(1/4) e^(i((5π)/8)) =z_1 (doubles) I = ∫_(−∞) ^(+∞) ((cos(πx))/((x^2 +1+i)^2 ))dx =Re( ∫_(−∞) ^(+∞) (e^(iπx) /((x^2 +1+i)^2 ))dx) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_0 ) Res(ϕ,z_0 )=lim_(z→z_0 ) (1/((2−1)!)) ((z−z_0 )^2 ϕ(z))^′ =lim_(z→z_0 ) ( (e^(iπz) /((z −z_1 )^2 )))^′ =lim_(z→z_0 ) ((iπ e^(iπz) (z−z_1 )^2 −2(z−z_1 )e^(iπz) )/((z−z_1 )^4 )) =lim_(z→z_0 ) ((iπ e^(iπz) (z−z_1 ) −2 e^(iπz) )/((x−z_1 )^3 )) = ((iπ e^(iπz_0 ) (z_0 −z_1 ) −2 e^(iπz_0 ) )/((z_0 −z_1 )3)) ∫_(−∞) ^(+∞) ϕ(z)dz =(−2π (z_0 −z_1 −2)(cos(πz_0 ) +isin(πz_0 ))(z−z_0 )^(−3) I =Re ( ∫_(−∞) ^(+∞) ϕ(z)dz) .

$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{i}\pi{z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}+{i}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:{z}^{\mathrm{2}} +\mathrm{1}+{i}\:={z}^{\mathrm{2}} \:−\left(−\mathrm{1}−{i}\right) \\ $$$$−\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}\:\left(\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:=\sqrt{\mathrm{2}}\left({cos}\left(\pi+\frac{\pi}{\mathrm{4}}\right)\:+{i}\:{sin}\left(\pi+\frac{\pi}{\mathrm{4}}\right)\right) \\ $$$$=\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \Rightarrow\:{z}^{\mathrm{2}} −\left(−\mathrm{1}−{i}\right)\:={z}^{\mathrm{2}} \:\:−\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \\ $$$$=\left({z}\:−^{\mathrm{4}} \left(\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{8}}} \right)^{\mathrm{2}} \:\:{so}\:{the}\:{poles}\:{are}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{8}}} ={z}_{\mathrm{0}} \right. \\ $$$${and}\:−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{8}}} ={z}_{\mathrm{1}} \:\:\:\left({doubles}\right) \\ $$$${I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}+{i}\right)^{\mathrm{2}} }{dx}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\pi{x}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}+{i}\right)^{\mathrm{2}} }{dx}\right) \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{0}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{0}} \right)={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left(\left({z}−{z}_{\mathrm{0}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right)^{'} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{0}} \:\:} \:\:\:\left(\:\frac{{e}^{{i}\pi{z}} }{\left({z}\:−{z}_{\mathrm{1}} \right)^{\mathrm{2}} }\right)^{'} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\frac{{i}\pi\:{e}^{{i}\pi{z}} \left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \:−\mathrm{2}\left({z}−{z}_{\mathrm{1}} \right){e}^{{i}\pi{z}} }{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{0}} } \:\:\frac{{i}\pi\:{e}^{{i}\pi{z}} \left({z}−{z}_{\mathrm{1}} \right)\:\:−\mathrm{2}\:{e}^{{i}\pi{z}} }{\left({x}−{z}_{\mathrm{1}} \right)^{\mathrm{3}} } \\ $$$$=\:\frac{{i}\pi\:{e}^{{i}\pi{z}_{\mathrm{0}} } \left({z}_{\mathrm{0}} \:−{z}_{\mathrm{1}} \right)\:−\mathrm{2}\:{e}^{{i}\pi{z}_{\mathrm{0}} } }{\left({z}_{\mathrm{0}} \:−{z}_{\mathrm{1}} \right)\mathrm{3}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\left(−\mathrm{2}\pi\:\left({z}_{\mathrm{0}} \:−{z}_{\mathrm{1}} \:−\mathrm{2}\right)\left({cos}\left(\pi{z}_{\mathrm{0}} \right)\:+{isin}\left(\pi{z}_{\mathrm{0}} \right)\right)\left({z}−{z}_{\mathrm{0}} \right)^{−\mathrm{3}} \right. \\ $$$${I}\:={Re}\:\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:. \\ $$

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