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Question Number 143083 by Mathspace last updated on 09/Jun/21

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

Answered by Olaf_Thorendsen last updated on 10/Jun/21

$$f_b\left(a\right)=\mathrm{\Psi }(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$ $$f_b^{\prime }\left(a\right)=\frac{\partial \mathrm{\Psi }}{\partial a}(a,b)=-{\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$ $$f_b^{″}\left(a\right)=\frac{{\partial }^2\mathrm{\Psi }}{\partial a^2}(a,b)=+{\int }_0^{\mathrm{\infty }}\frac{x^4{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$ $$$$ $$f_b^{″}\left(a\right)-2b^2{f}_b^{\prime }\left(a\right)+b^4{f}_b\left(a\right)$$ $$={\int }_0^{\mathrm{\infty }}\frac{(x^4+2b^2{x}^2+b^4)e^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$ $$={\int }_0^{\mathrm{\infty }}\frac{{(x^2+b^2)}^2{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$ $$={\int }_0^{\mathrm{\infty }}{e}^{-{ax}^2}dx$$ $$=\frac{1}{\sqrt{a}}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt$$ $$=\frac{1}{2}\sqrt{\frac{\pi }{a}}\left(\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt\right)$$ $$=\frac{1}{2}\sqrt{\frac{\pi }{a}}$$ $$$$ $$f_b^{″}\left(a\right)-2b^2{f}_b^{\prime }\left(a\right)+b^4{f}_b\left(a\right)=\frac{1}{2}\sqrt{\frac{\pi }{a}}$$ $$$$ $$f_b\left(a\right)=\mathrm{\Psi }(a,b)=$$ $$(c_1+c_{2}a)e^{{ab}^2}+\frac{\sqrt{\pi }}{2}{e}^{{ab}^2}[a\mathrm{\Gamma }(\frac{1}{2},b^{2}a)+\frac{\mathrm{\Gamma }(\frac{3}{2},{ab}^2)}{b^3}]$$ $$...\mathrm{to}\mathrm{be}\mathrm{continued}...$$

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