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Question Number 174727 by princeDera last updated on 09/Aug/22

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

Answered by aleks041103 last updated on 10/Aug/22

$${\int }_0^{\mathrm{\infty }}\frac{1}{x^2+a^2}\frac{e^{-x^2}}{x^2+a^2}dx=$$$$=\frac{1}{a^2}\int \frac{(x^2+a^2-x^2)}{x^2+a^2}\frac{e^{-x^2}}{x^2+a^2}dx=$$$$=\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{a^2+x^2}dx+\frac{1}{a^2}{\int }_0^{\mathrm{\infty }}\frac{-x^2{e}^{-x^2}}{{(x^2+a^2)}^2}dx$$$${\int }_0^{\mathrm{\infty }}\frac{-x^2{e}^{-x^2}}{{(x^2+a^2)}^2}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{xe}^{-x^2}(-\frac{2xdx}{{(x^2+a^2)}^2})=$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{xe}^{-x^2}d\left(\frac{1}{x^2+a^2}\right)=$$$$={\left[\frac{{xe}^{-x^2}}{2(x^2+a^2)}\right]}_0^{\mathrm{\infty }}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{x^2+a^2}(e^{-x^2}-2x^2{e}^{-x^2})dx=$$$$={\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-x^2}}{x^2+a^2}dx-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{x^2+a^2}dx=$$$$={\int }_0^{\mathrm{\infty }}(1-\frac{a^2}{x^2+a^2})e^{-x^2}dx-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{x^2+a^2}dx=$$$$={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx-(\frac{1}{2}+a^2){\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}dx}{x^2+a^2}=$$$$=\frac{\sqrt{\pi }}{2}-(\frac{1}{2}+a^2){\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}dx}{x^2+a^2}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+a^2)}^2}dx=\frac{\sqrt{\pi }}{2a^2}+(\frac{1}{2a^2}-1){\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}dx}{x^2+a^2}$$

Commented by aleks041103 last updated on 10/Aug/22

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}dx}{x^2+a^2}=\frac{1}{a}{\int }_0^{\mathrm{\infty }}\frac{e^{-a^2{\left(x/a\right)}^2}d\left(x/a\right)}{{\left(x/a\right)}^2+1}=\frac{1}{a}f\left(a\right)$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-a^2{x}^2}dx}{x^2+1}$$$$f{}^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{-2{ax}^2{e}^{-a^2{x}^2}}{x^2+1}dx=2a{\int }_0^{\mathrm{\infty }}(\frac{1}{x^2+1}-1)e^{-{\left(ax\right)}^2}dx=$$$$=2a{\int }_0^{\mathrm{\infty }}\frac{e^{-a^2{x}^2}dx}{x^2+1}-2{\int }_0^{\mathrm{\infty }}{e}^{-{\left(ax\right)}^2}d\left(ax\right)=$$$$=2af\left(a\right)-\sqrt{\pi }$$$$\Rightarrow f{}^{\prime }\left(a\right)=2af\left(a\right)-\sqrt{\pi }$$$$\Rightarrow e^{-a^2}f{}^{\prime }-2{ae}^{-a^2}f=-\sqrt{\pi }{e}^{-a^2}$$$$\left(e^{-a^2}\right)f{}^{\prime }+{\left(e^{-a^2}\right)}^{\prime }f={\left(e^{-a^2}f\right)}^{\prime }=-\sqrt{\pi }{e}^{-a^2}$$$$\Rightarrow e^{-a^2}f\left(a\right)-0=-\sqrt{\pi }{\int }_{\mathrm{\infty }}^a{e}^{-x^2}dx=+\sqrt{\pi }(q\left(\mathrm{\infty }\right)-q\left(a\right))$$$$\Rightarrow q\left(a\right)=\frac{\sqrt{\pi }}{2}erf\left(a\right)$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+a^2)}^2}dx=\frac{\sqrt{\pi }}{2a^2}+(\frac{1}{2a^2}-1)\frac{1}{a}f\left(a\right)$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+a^2)}^2}dx=\frac{\sqrt{\pi }}{2a^2}-\frac{\pi }{2a}(1-\frac{1}{2a^2})e^{a^2}(1-erf\left(a\right))$$$$thisistruefora>0$$$$buttheinitialintegralisanevenfunctionina$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+a^2)}^2}dx=\frac{\sqrt{\pi }}{2a^2}-\frac{\pi }{2\mid a\mid }(1-\frac{1}{2a^2})e^{a^2}(1-erf(\mid a\mid ))$$

Commented by princeDera last updated on 10/Aug/22

$$thankyousir$$

Commented by peter frank last updated on 11/Aug/22

$$\mathrm{thank}\mathrm{you}$$

Commented by Tawa11 last updated on 11/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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