TITLE: How to evaluate this monsterous integral?
QUESTION [9 upvotes]: How to prove this?
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  \frac {2(x-x^2)e^{-(x^2+y^2+z^2)}}{((x-1)^2+(y-1)^2+(z-1)^2)^{3/2}}dxdydz=\frac {-4\pi e^{-3}}{3}$$
I tried converting to polar coordinates but still I couldn't evaluate.

REPLY [3 votes]: Another answer:
I noticed it also follows from this identity see identity
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  \frac {\partial f(x',y',z')}{\partial x'}\frac {x'-x}{4\pi r'^3}dx'dy'dz'+\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  \frac {\partial f(x',y',z')}{\partial y'}\frac {y'-y}{4\pi r'^3}dx'dy'dz'+
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  \frac {\partial f(x',y',z')}{\partial z'}\frac {z'-z}{4\pi r'^3}dx'dy'dz'=-f(x,y,z)$$
with $$f(x,y,z)=e^{-(x^2+y^2+z^2)}$$
it is easily seen when $x=y=z=1$ $\quad$ $f(1,1,1)=e^{-3}$
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}  \frac {\partial f(x',y',z')}{\partial x'}\frac {x'-1}{4\pi r'^3}dx'dy'dz'=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac {2(x'-x'^2)e^{-(x'^2+y'^2+z'^2)}}{4\pi((x'-1)^2+(y'-1)^2+(z'-1)^2)^{3/2}}dx'dy'dz'=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac {2(y'-y'^2)e^{-(x'^2+y'^2+z'^2)}}{4\pi((x'-1)^2+(y'-1)^2+(z'-1)^2)^{3/2}}dx'dy'dz'=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac {2(z'-z'^2)e^{-(x'^2+y'^2+z'^2)}}{4\pi((x'-1)^2+(y'-1)^2+(z'-1)^2)^{3/2}}dx'dy'dz'$$
Therefore: $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac {2(x'-x'^2)e^{-(x'^2+y'^2+z'^2)}}{4\pi((x'-1)^2+(y'-1)^2+(z'-1)^2)^{3/2}}dx'dy'dz'=\frac {-f(1,1,1)}{3}$$
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac {2(x'-x'^2)e^{-(x'^2+y'^2+z'^2)}}{((x'-1)^2+(y'-1)^2+(z'-1)^2)^{3/2}}dx'dy'dz'=\frac {-4\pi e^{-3}}{3}$$