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Question Number 59991 by selimatas01 last updated on 16/May/19

Commented by Mr X pcx last updated on 16/May/19

let I =∫_0 ^∞ ((cosx)/((x^2 +1)^2 )) dx ⇒ 2I =∫_(−∞) ^(+∞) ((cosx)/((x^2 +1)^2 )) dx= Re( ∫_(−∞) ^(+∞) (e^(ix) /((x^2 +1)^2 )) dx) let consider the complexe function ϕ(z) =(e^(iz) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(iz) /((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and −i(doubles) redidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i) Res(ϕ,i)=lim_(z→i) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(e^(iz) /((z+i)^2 ))}^((1)) =lim_(z→i) ((i e^(iz) (z+i)^2 −2(z+i)e^(iz) )/((z+i)^4 )) =lim_(z→i) ((ie^(iz) (z+i)−2e^(iz) )/((z+i)^3 )) =((−2 e^(−1) −2e^(−1) )/((2i)^3 )) =((−4 e^(−1) )/(−8i)) =(e^(−1) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(e^(−1) /(2i)) =πe^(−1) ⇒ 2I =(π/e) ⇒★I =(π/(2e)) ★

$${let}\:\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cosx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cosx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}= \\ $$$${Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{ix}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\right)\:{let}\:{consider} \\ $$$${the}\:{complexe}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{{e}^{{iz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{iz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\left({doubles}\right) \\ $$$${redidus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left\{\frac{{e}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{{i}\:{e}^{{iz}} \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{{ie}^{{iz}} \left({z}+{i}\right)−\mathrm{2}{e}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}\:{e}^{−\mathrm{1}} −\mathrm{2}{e}^{−\mathrm{1}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{4}\:{e}^{−\mathrm{1}} }{−\mathrm{8}{i}}\:=\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\frac{{e}^{−\mathrm{1}} }{\mathrm{2}{i}}\:=\pi{e}^{−\mathrm{1}} \:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\frac{\pi}{{e}}\:\Rightarrow\bigstar{I}\:=\frac{\pi}{\mathrm{2}{e}}\:\bigstar \\ $$

Commented by selimatas01 last updated on 17/May/19

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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