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Question Number 143082 by Mathspace last updated on 09/Jun/21

$$calculatef(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-{ax}^2}}{x^2+b^2}dx$$$$witha>0andb>0$$

Answered by Dwaipayan Shikari last updated on 09/Jun/21

$$={\int }_0^{\mathrm{\infty }}{e}^{-{ax}^2}{\int }_0^{\mathrm{\infty }}{e}^{-t(x^2+b)}dxdt$$$$=\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-tb}}{\sqrt{t+a}}dtt+a=u^2$$$$=\sqrt{\pi }{e}^{ab}{\int }_{\sqrt{a}}^{\mathrm{\infty }}{e}^{-u^2}du$$$$=\sqrt{\pi }{e}^{ab}(\frac{\sqrt{\pi }}{2}-\frac{\sqrt{\pi }}{2}erf\left(\sqrt{a}\right))=\frac{\pi }{2}{e}^{ab}\left(erfc\left(\sqrt{a}\right)\right)$$

Answered by mathmax by abdo last updated on 10/Jun/21

$$\mathrm{f}(\mathrm{a},\mathrm{b})={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{ax}}^2}}{{\mathrm{x}}^2+{\mathrm{b}}^2}\mathrm{dx}={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{x}}^2+{\mathrm{b}}^2)\mathrm{t}}\mathrm{dt}\right){\mathrm{e}}^{-{\mathrm{ax}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{t}+\mathrm{a}){\mathrm{x}}^2}\mathrm{dx}\right){\mathrm{e}}^{-{\mathrm{b}}^2\mathrm{t}}\mathrm{dt}[\mathrm{but}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{t}+\mathrm{a}){\mathrm{x}}^2}\mathrm{dx}{=}_{\sqrt{\mathrm{t}+\mathrm{a}}\mathrm{x}=\mathrm{y}}{\int }_{\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\frac{\mathrm{dy}}{\sqrt{\mathrm{t}+\mathrm{a}}}\Rightarrow$$$$\mathrm{f}(\mathrm{a},\mathrm{b})={\int }_{\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy}.{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{b}}^2\mathrm{t}}}{\sqrt{\mathrm{t}+\mathrm{a}}}\mathrm{dt}(\mathrm{let}{\lambda }_0={\int }_{\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2}\mathrm{dy})$$$${=}_{\mathrm{t}+\mathrm{a}={\mathrm{z}}^2}{\lambda }_0{\int }_{\sqrt{\mathrm{a}}}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{b}}^2({\mathrm{z}}^2-\mathrm{a})}}{\mathrm{z}}\left(2\mathrm{z}\right)\mathrm{dz}=2{\lambda }_0{\mathrm{e}}^{{\mathrm{ab}}^2}{\int }_{\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{b}}^2{\mathrm{z}}^2}\mathrm{dz}$$$${=}_{\mathrm{bz}=\mathrm{u}}2{\lambda }_0{\mathrm{e}}^{{\mathrm{ab}}^2}{\int }_{\mathrm{b}\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\frac{\mathrm{du}}{\mathrm{b}}$$$$=\frac{2{\lambda }_0}{\mathrm{b}}{\mathrm{e}}^{{\mathrm{ab}}^2}{\int }_{\mathrm{b}\sqrt{\mathrm{a}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}$$

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