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Question Number 142989 by mathmax by abdo last updated on 08/Jun/21

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{3\mathrm{x}}^2}}{1+{\mathrm{x}}^2}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 08/Jun/21

$${\int }_0^{\mathrm{\infty }}{e}^{-3x^2}{\int }_0^{\mathrm{\infty }}{e}^{-u(1+x^2)}dxdu$$$$=\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-u}}{\sqrt{u+3}}duu+3=t^2$$$$=e^3\sqrt{\pi }{\int }_{\sqrt{3}}^{\mathrm{\infty }}{e}^{-t^2}dt$$$$=e^3\sqrt{\pi }({\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt-{\int }_0^{\sqrt{3}}{e}^{-t^2}dt)$$$$Now\frac{2}{\sqrt{\pi }}{\int }_0^x{e}^{-t^2}dx=efr\left(x\right)$$$$=e^3\frac{\pi }{2}-e^3\frac{erf\left(\sqrt{3}\right)}{2}\pi =\frac{e^3\pi }{2}(1-erf\left(\sqrt{3}\right))$$

Answered by mathmax by abdo last updated on 09/Jun/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{3\mathrm{x}}^2}}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}(1+{\mathrm{x}}^2)}\mathrm{dt}\right){\mathrm{e}}^{-{3\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{t}+3){\mathrm{x}}^2}\mathrm{dx}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mathrm{but}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{t}+3){\mathrm{x}}^2}\mathrm{dx}{=}_{\sqrt{\mathrm{t}+3}\mathrm{x}=\mathrm{y}}{\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\frac{\mathrm{dy}}{\sqrt{\mathrm{t}+3}}=\frac{1}{\sqrt{\mathrm{t}+3}}{\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}\Rightarrow$$$$\mathrm{\Phi }={\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{\sqrt{\mathrm{t}+3}}\mathrm{dt}\mathrm{and}$$$${\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{\sqrt{\mathrm{t}+3}}\mathrm{dt}{=}_{\sqrt{\mathrm{t}+3}=\mathrm{z}}{\int }_{\sqrt{3}}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-({\mathrm{z}}^2-3)}}{\mathrm{z}}\left(2\mathrm{z}\right)\mathrm{dz}={2\mathrm{e}}^3{\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}\Rightarrow$$$$\mathrm{\Phi }={2\mathrm{e}}^3{\left({\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}\right)}^2\mathrm{let}{\lambda }_0={\int }_{\sqrt{3}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}\Rightarrow$$$$\mathrm{\Phi }={2\mathrm{e}}^3{\lambda }_0^2$$

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