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Question Number 27802 by abdo imad last updated on 15/Jan/18

find the value of ∫_0 ^∞ (e^(−2x^2 ) /((3+x^2 )^2 ))dx .

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:. \\ $$

Commented by abdo imad last updated on 17/Jan/18

let put I= ∫_0 ^∞ (e^(−2x^2 ) /((3+x^2 )^2 ))dx I= (1/2) ∫_(−∝) ^(+∝) (e^(−2x^2 ) /((3+x^2 )^2 )) dx let introduce the complex function f(z)= (e^(−2z^2 ) /((3+z^2 )^2 )) .poles of f ? f(z)= (e^(−2z^2 ) /((z −i(√3))^2 (z+i(√3))^2 )) so f have a double poles i(√3) and −i(√3) ∫_(−∝) ^(+∝) f(z)dz =2iπ Res (f,i(√3)) Res(f,i(√(3)))= lim_(z→i(√3) ) (1/((2−1)!)) ((z−i(√3))^2 f(z))^′ lim_(z→i(√3)) ( e^(−2z^2 ) .(z+i(√3))^(−2) )^, =lim_(z→i(√3)) ( −4z e^(−z^2 ) (z+i(√3))^(−2) −2 e^(−2z^2 ) (z+i(√3))^(−3) ) = −4(i(√3)) e^3 (2i(√3))^(−2) −2 e^6 (2i(√3))^(−3) = ((−4i(√3) e^3 )/(−12)) − ((2 e^6 )/(−24i(√3)))= ((i(√3) e^3 )/3) + (e^6 /(12i(√3))). ∫_(−∝) ^(+∝) f(z)dz =2iπ( ((i(√3) e^3 )/3) + (e^6 /(12i(√3))) ) = ((−2(√3) e^3 )/3) + (e^6 /(6(√3))) finally I= (1/2) ∫_R f(z)dz= ((−e^3 (√3))/3) + (e^6 /(12(√3))) .

$${let}\:{put}\:\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\propto} ^{+\propto} \:\:\:\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:\:{let}\:{introduce}\:{the}\:{complex}\:{function} \\ $$$${f}\left({z}\right)=\:\:\frac{{e}^{−\mathrm{2}{z}^{\mathrm{2}} } }{\left(\mathrm{3}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:.{poles}\:{of}\:{f}\:\:? \\ $$$${f}\left({z}\right)=\:\:\:\:\:\frac{{e}^{−\mathrm{2}{z}^{\mathrm{2}} } }{\left({z}\:−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:\:{so}\:{f}\:{have}\:{a}\:{double}\:{poles}\:{i}\sqrt{\mathrm{3}} \\ $$$${and}\:−{i}\sqrt{\mathrm{3}} \\ $$$$\int_{−\propto} ^{+\propto} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\:\left({f},{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left({f},{i}\sqrt{\left.\mathrm{3}\right)}=\:{lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}\:} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left(\left({z}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {f}\left({z}\right)\right)^{'} \right. \\ $$$${lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\left(\:{e}^{−\mathrm{2}{z}^{\mathrm{2}} } .\left({z}+{i}\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \right)^{,} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\left(\:\:−\mathrm{4}{z}\:{e}^{−{z}^{\mathrm{2}} } \left({z}+{i}\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \:−\mathrm{2}\:{e}^{−\mathrm{2}{z}^{\mathrm{2}} } \left({z}+{i}\sqrt{\mathrm{3}}\right)^{−\mathrm{3}} \right) \\ $$$$=\:−\mathrm{4}\left({i}\sqrt{\mathrm{3}}\right)\:{e}^{\mathrm{3}} \:\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \:\:−\mathrm{2}\:{e}^{\mathrm{6}} \:\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)^{−\mathrm{3}} \\ $$$$=\:\:\frac{−\mathrm{4}{i}\sqrt{\mathrm{3}}\:{e}^{\mathrm{3}} }{−\mathrm{12}}\:\:−\:\frac{\mathrm{2}\:{e}^{\mathrm{6}} }{−\mathrm{24}{i}\sqrt{\mathrm{3}}}=\:\frac{{i}\sqrt{\mathrm{3}}\:{e}^{\mathrm{3}} }{\mathrm{3}}\:\:+\:\frac{{e}^{\mathrm{6}} }{\mathrm{12}{i}\sqrt{\mathrm{3}}}. \\ $$$$\int_{−\propto} ^{+\propto} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\:\frac{{i}\sqrt{\mathrm{3}}\:{e}^{\mathrm{3}} }{\mathrm{3}}\:\:+\:\:\frac{{e}^{\mathrm{6}} }{\mathrm{12}{i}\sqrt{\mathrm{3}}}\:\right) \\ $$$$=\:\:\frac{−\mathrm{2}\sqrt{\mathrm{3}}\:{e}^{\mathrm{3}} }{\mathrm{3}}\:\:+\:\:\:\frac{{e}^{\mathrm{6}} }{\mathrm{6}\sqrt{\mathrm{3}}}\:\:{finally} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} {f}\left({z}\right){dz}=\:\frac{−{e}^{\mathrm{3}} \sqrt{\mathrm{3}}}{\mathrm{3}}\:\:+\:\:\frac{{e}^{\mathrm{6}} }{\mathrm{12}\sqrt{\mathrm{3}}}\:\:. \\ $$

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