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Question Number 147203 by mathmax by abdo last updated on 18/Jul/21

$$\mathrm{find}{\mathrm{U}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{nx}}^2}}{{\mathrm{x}}^2+{\mathrm{n}}^2}\mathrm{dx}(\mathrm{n}⩾1)$$$$\mathrm{nature}\mathrm{of}\mathrm{\Sigma }{\mathrm{U}}_{\mathrm{n}}\mathrm{and}\mathrm{\Sigma }{\mathrm{nU}}_{\mathrm{n}}$$

Answered by mathmax by abdo last updated on 23/Jul/21

$${\mathrm{U}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{nx}}^2}}{{\mathrm{x}}^2+{\mathrm{n}}^2}\mathrm{dx}{=}_{\mathrm{x}=\mathrm{nt}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{n}}^3{\mathrm{t}}^2}}{{\mathrm{n}}^2({\mathrm{t}}^2+1)}\mathrm{dt}$$$$=\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{t}}^2+1)\mathrm{x}}\mathrm{dx}\right){\mathrm{e}}^{-{\mathrm{n}}^3{\mathrm{t}}^2}\mathrm{dt}$$$$=\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{x}+{\mathrm{n}}^3){\mathrm{t}}^2}\mathrm{dt}\right){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}$$$${=}_{\sqrt{\mathrm{x}+{\mathrm{n}}^3}\mathrm{t}=\mathrm{z}}\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\left({\int }_{\sqrt{\mathrm{x}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\frac{\mathrm{dz}}{\sqrt{\mathrm{x}+{\mathrm{n}}^3}}\right){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{{\mathrm{n}}^2}{\int }_0^{\mathrm{\infty }}\left(\frac{{\lambda }_0\mathrm{erf}\left(\sqrt{\mathrm{x}}\right)}{\sqrt{\mathrm{x}+{\mathrm{n}}^3}}\right){\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}...\mathrm{be}\mathrm{continued}...$$

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