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Question Number 200801 by mnjuly1970 last updated on 23/Nov/23

Answered by witcher3 last updated on 24/Nov/23

$$\mathrm{introduce}\mathrm{erfc}\left(\mathrm{x}\right)=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}$$$$\varphi ={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{(\mathrm{x}+\mathrm{y}+\mathrm{z})}^2}\mathrm{dxdydz}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{(\mathrm{x}+\mathrm{y}+\mathrm{z})}^2}\mathrm{dx},\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{r}$$$$={\int }_{\mathrm{y}+\mathrm{z}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{dr}=\frac{\sqrt{\pi }}{2}-{\int }_0^{\mathrm{y}+\mathrm{z}}{\mathrm{e}}^{-{\mathrm{r}}^2}\mathrm{dr}=\frac{\sqrt{\pi }}{2}(1-\mathrm{erfc}(\mathrm{y}+z)$$$$\varphi =\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}(1-\mathrm{erfc}(\mathrm{y}+\mathrm{z}))\mathrm{dydz}=\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{z}}^{\mathrm{\infty }}(1-\mathrm{erfc}\left(\mathrm{y}\right))\mathrm{dydz}$$$${\int }_{\mathrm{z}}^{\mathrm{\infty }}(1-\mathrm{erfc}\left(\mathrm{y}\right)\mathrm{dy}=\underset{x\to \mathrm{\infty }}{\lim }{\int }_{\mathrm{z}}^{\mathrm{x}}(1-\mathrm{erfc}\left(\mathrm{y}\right))\mathrm{dy}$$$$=\underset{x\to \mathrm{\infty }}{\lim }{[\mathrm{y}-\mathrm{yerfc}\left(\mathrm{y}\right)]}_{\mathrm{z}}^{\mathrm{x}}+\frac{2}{\sqrt{\pi }}{\int }_{\mathrm{z}}^{\mathrm{x}}{\mathrm{ye}}^{-{\mathrm{y}}^2}\mathrm{dy}$$$$=\underset{x\to \mathrm{\infty }}{\lim }(\mathrm{x}-\mathrm{xerfc}\left(\mathrm{x}\right)-\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{\sqrt{\pi }}+\mathrm{zerfc}\left(\mathrm{z}\right)-\mathrm{z}+\frac{{\mathrm{e}}^{-{\mathrm{z}}^2}}{\sqrt{\pi }})$$$$\underset{x\to \mathrm{\infty }}{\lim }(\mathrm{x}-\mathrm{xerfc}\left(\mathrm{x}\right))=\underset{x\to \mathrm{\infty }}{\lim }(\mathrm{x}-\frac{2\mathrm{x}}{\sqrt{\pi }}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt})$$$$=\underset{x\to \mathrm{\infty }}{\lim }\mid \mathrm{x}-\mathrm{x}(1-\frac{2}{\sqrt{\pi }}{\int }_{\mathrm{x}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt})\mid ⩽\underset{x\to \mathrm{\infty }}{\lim }\mid \frac{2\mathrm{x}}{\sqrt{\pi }}{\int }_{\mathrm{x}}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\mid$$$$=\underset{x\to \mathrm{\infty }}{\lim }\left({\mathrm{xe}}^{-\mathrm{x}}\right)=0$$$$\mathrm{\varnothing }=\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}(\mathrm{zerfc}\left(\mathrm{z}\right)-\mathrm{z}+\frac{{\mathrm{e}}^{-{\mathrm{z}}^2}}{\sqrt{\pi }})\mathrm{dz}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}=\frac{\sqrt{\pi }}{2}$$$${\int }_0^{\mathrm{\infty }}\mathrm{z}(\mathrm{erfc}\left(\mathrm{z}\right)-1)\mathrm{dz}={\left[\frac{{\mathrm{z}}^2}{2}(\mathrm{erfc}\left(\mathrm{z}\right)-1)\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{z}}^2}{2}\left(\frac{2}{\sqrt{\pi }}{\mathrm{e}}^{-{\mathrm{z}}^2}\right)\mathrm{dz}$$$$=-\frac{1}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^2{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}=-\frac{1}{2\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{3}{2}-1}{\mathrm{e}}^{-\mathrm{t}}=-\frac{1}{4}$$$$\varphi =\frac{\sqrt{\pi }}{2}(-\frac{1}{4}+\frac{1}{2})=\frac{\sqrt{\pi }}{8}$$$$\underset{x\to \mathrm{\infty }}{\lim }(\mid \frac{{\mathrm{x}}^2}{2}(\mathrm{erfc}\left(\mathrm{x}\right)-1)\mid )=\frac{{\mathbf{x}}^2}{2}{\int }_{\mathrm{x}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}⩽\frac{{\mathrm{x}}^2}{2}{\mathrm{e}}^{-\mathrm{x}}\to 0$$$$\varphi =\frac{\sqrt{\pi }}{8}$$$$$$

Commented by mnjuly1970 last updated on 24/Nov/23

$$thanksalotsirsonicesolution$$

Commented by witcher3 last updated on 24/Nov/23

$$\mathrm{withe}\mathrm{Pleasur}$$

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