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

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{e^{-2x^2}}{{(3+x^2)}^2}dx.$$

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

$$letputI={\int }_0^{\mathrm{\infty }}\frac{e^{-2x^2}}{{(3+x^2)}^2}dx$$$$I=\frac{1}{2}{\int }_{-\propto }^{+\propto }\frac{e^{-2x^2}}{{(3+x^2)}^2}dxletintroducethecomplexfunction$$$$f\left(z\right)=\frac{e^{-2z^2}}{{(3+z^2)}^2}.polesoff?$$$$f\left(z\right)=\frac{e^{-2z^2}}{{(z-i\sqrt{3})}^2{(z+i\sqrt{3})}^2}sofhaveadoublepolesi\sqrt{3}$$$$and-i\sqrt{3}$$$${\int }_{-\propto }^{+\propto }f\left(z\right)dz=2i\pi Res(f,i\sqrt{3})$$$$Res(f,i\sqrt{3)}={lim}_{z\to i\sqrt{3}}\frac{1}{(2-1)!}{\left({(z-i\sqrt{3})}^{2}f\left(z\right)\right)}^{{}^{\prime }}$$$${lim}_{z\to i\sqrt{3}}{(e^{-2z^2}.{(z+i\sqrt{3})}^{-2})}^{,}$$$$={lim}_{z\to i\sqrt{3}}(-4z{e}^{-z^2}{(z+i\sqrt{3})}^{-2}-2e^{-2z^2}{(z+i\sqrt{3})}^{-3})$$$$=-4\left(i\sqrt{3}\right)e^3{\left(2i\sqrt{3}\right)}^{-2}-2e^6{\left(2i\sqrt{3}\right)}^{-3}$$$$=\frac{-4i\sqrt{3}{e}^3}{-12}-\frac{2e^6}{-24i\sqrt{3}}=\frac{i\sqrt{3}{e}^3}{3}+\frac{e^6}{12i\sqrt{3}}.$$$${\int }_{-\propto }^{+\propto }f\left(z\right)dz=2i\pi (\frac{i\sqrt{3}{e}^3}{3}+\frac{e^6}{12i\sqrt{3}})$$$$=\frac{-2\sqrt{3}{e}^3}{3}+\frac{e^6}{6\sqrt{3}}finally$$$$I=\frac{1}{2}{\int }_{R}f\left(z\right)dz=\frac{-e^3\sqrt{3}}{3}+\frac{e^6}{12\sqrt{3}}.$$

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