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Question Number 49967 by maxmathsup by imad last updated on 12/Dec/18
calculate ∫_0 ^(+∞) (dx/((x^2 −i)^2 ))
$$\:\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} } \\ $$
Commented by Abdo msup. last updated on 13/Dec/18
let A =∫_0 ^∞ (dx/((x^2 −i)^2 )) ⇒ A =(1/2)∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 )) let consider the complex function ϕ(z)=(1/((z^2 −i)^2 )) we have ϕ(z) = (1/((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 )) so the poles of ϕ are +^− e^((iπ)/4) (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/4) ) but Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)) { (z−e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/4) ) { (1/((z+e^((iπ)/4) )^2 ))}^((1)) =lim_(z→e^((iπ)/4) ) −2(((z+e^((iπ)/4) ))/((z+e^((iπ)/4) )^4 )) =lim_(z→ e^((iπ)/4) ) ((−2)/((z+e^((iπ)/4) )^3 )) =((−2)/((2e^((iπ)/4) )^3 )) =((−1)/(4 e^(i((3π)/4)) )) =((−1)/(4 e^(i(π−(π/4))) )) =(1/4) e^((iπ)/4) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(1/4) e^((iπ)/4) =((iπ)/2)( (1/( (√2))) +(i/( (√2)))) =((iπ)/(2(√2))) −(π/(2(√2))) = 2A . remark we have ∫_(−∞) ^∞ (dx/((x^2 −i)^2 )) =∫_(−∞) ^∞ (((x^2 +i)^2 )/((x^4 +1)^2 )) dx =∫_(−∞) ^∞ ((x^4 +2ix^2 −1)/((x^4 +1)^2 )) dx =−(π/(2(√2))) +i(π/(2(√2))) ∫_(−∞) ^∞ ((x^4 −1)/((x^4 +1)^2 )) dx =−(π/(2(√2))) and ∫_(−∞) ^∞ ((2x^2 )/((x^4 +1)^2 )) dx =(π/(2(√2))) .
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:\Rightarrow\:{A}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:\:{let}\:{consider} \\ $$$${the}\:{complex}\:{function}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} }\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\left({doubles}\right)\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:\:{but} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\:\left\{\:\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\left\{\:\frac{\mathrm{1}}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:−\mathrm{2}\frac{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow\:{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\:\:\frac{−\mathrm{2}}{\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{\left(\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }\:=\frac{−\mathrm{1}}{\mathrm{4}\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} }\:\:=\frac{−\mathrm{1}}{\mathrm{4}\:{e}^{{i}\left(\pi−\frac{\pi}{\mathrm{4}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:=\frac{{i}\pi}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\frac{{i}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\mathrm{2}{A}\:. \\ $$$${remark}\:{we}\:{have}\:\int_{−\infty} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} −{i}\right)^{\mathrm{2}} } \\ $$$$=\int_{−\infty} ^{\infty} \:\:\frac{\left({x}^{\mathrm{2}} \:+{i}\right)^{\mathrm{2}} }{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=\int_{−\infty} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} \:+\mathrm{2}{ix}^{\mathrm{2}} \:−\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:+{i}\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{\infty} \:\:\:\frac{{x}^{\mathrm{4}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:{and}\:\int_{−\infty} ^{\infty} \:\:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$$$ \\ $$$$ \\ $$

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