Here is what i'm given:
"The so called Schwinger parametrization is based on identities like the following:
$$\frac{1}{\sqrt{R}}=\frac{2}{\sqrt{\pi}}\int_0^{\infty}e^{-Ru^2}du$$
The first octant of a three-dimensional coordinate system is filled with continuously distributed charge of density $\rho(x,y,z)=\rho_0 e^{-(x^2+2y^2+2z^2)}$. Use the above relation to evaluate the electrostatic potential in the origin, which is given by the integral:
$$V_0=\frac{\rho_0}{4\pi \epsilon_o}\iiint_0^{\infty}\frac{e^{-(x^2+2y^2+2z^2)}}{\sqrt{x^2+y^2+z^2}} dxdydz$$
Hint: The Schwinger parametrization converts the triple integral into a quadrupple integral. Three of the four integrals are Gaussian and thus can be easily evaluated. Use a computer algebra system to evaluate the last one-dimensional integral that you obtain."
This is what i've tried (but i am probably completely off base): This problem looks like an obvious candidate for spherical coordinates so using the the identities $x=r\sin\phi\cos\theta$, $y=r\sin\phi\sin\theta$, $z=r\cos\phi$ i have tried to simplify the terms in the exponent as such:
$$Exp[-(x^2+2y^2+2z^2)]=Exp[-r^2(\sin^2\phi\cos^2\theta + 2\sin^2\phi\sin^2\theta + 2\cos^2\phi)]=Exp[-r^2(\frac{1-\cos2\phi}{2}\frac{1+\cos 2\theta}{2})+(1-\cos 2\phi)(1-\cos 2\theta)+(1+\cos 2\phi))]=Exp[-\frac{r^2}{4}(9-3\cos 2\theta -\cos 2\phi -3\cos 2\phi\cos 2\theta)]$$
Then using $r=\sqrt{x^2+y^2+z^2}$ and $dxdydz=r^2\sin\phi drd\phi d\theta$ i have rewritten the problem as:
$$V_0=\frac{\rho_0}{4\pi \epsilon_o}\int_0^{\infty}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}e^{-\frac{9r^2}{4}} e^{-\frac{3r^2}{4}\cos 2\theta} e^{-\frac{r^2}{4}\cos 2\phi} e^{-\frac{3r^2}{4}\cos 2\phi\cos 2\theta}r\sin\phi d\theta d\phi dr$$
Now this just doesn't seem right. I still have multiple variables in most of the exponents and i don't see how i could separate them. The problem may be that i haven't done any multivariable calculus in over a year but i think i a missing something pretty obvious. Any help would be much appreciated.
You're completely correct in noticing that spherical coordinates can simplify the equation. The only catch is the method of embedding such a sphere space. Here's what I mean. Originally, we are given the space: $$B=\left\{\begin{matrix}x=x,x=[0,\infty] \\ y=y, y=[0,\infty] \\ z=z, z=[0,\infty]\end{matrix}\right.$$ One noticible thing about the parametrization of the sphere you've given: $$\Omega_1=\left\{\begin{matrix}x=r\sin\phi\cos\theta \\ y=r\sin\phi\sin\theta \\ z=r\cos\phi\end{matrix},\phantom{..}\begin{matrix}r=[0,\infty] \\ \theta=[0..1/2\pi] \\ \phi=[0..1/2\pi]\end{matrix}\right.$$ Is that you'll find that: $$x^2+y^2=r^2\sin^2\phi\phantom{..}\&\phantom{..}x^2+y^2+z^2=r^2$$ By expanding the integral, we find that: $$V_0=\frac{\rho_0}{4\pi\epsilon_0}\int_B\frac{\text{e}^{-(x^2+2y^2+2z^2)}}{\sqrt{x^2+y^2+z^2}}dB=\frac{\rho_0}{4\pi\epsilon_0}\int_B\frac{\text{e}^{-(x^2+y^2+z^2)}\text{e}^{-(y^2+z^2)}}{\sqrt{x^2+y^2+z^2}}dB$$ We see that instead of having $x^2+y^2=r%2\sin^2\phi$, we need to have $z^2+y^2=r^2\sin^2\phi$. Thus we reformulate the space to the following parametrization: $$\Omega_2=\left\{\begin{matrix}x=r\cos\phi \\ y=r\sin\phi\sin\theta \\ z=r\sin\phi\cos\theta\end{matrix},\phantom{..}\begin{matrix}r=[0,\infty] \\ \theta=[0..1/2\pi] \\ \phi=[0..1/2\pi]\end{matrix}\right.$$ Since all the coordinates are equivalent, the angular parametrizations are equivalent. In addition, since this is describing a sphere, the volume element is the same. This produces the following integral: $$V_0=\frac{\rho_0}{4\pi\epsilon_0}\int_{\Omega_2}\frac{\text{e}^{-r^2}\text{e}^{-r^2\sin^2\phi}}{r}d\Omega_2=\frac{\rho_0}{4\pi\epsilon_0}\int_{\Omega_2}\frac{\text{e}^{-r^2(1+\sin^2\phi)}}{r}d\Omega_2$$ Implementing all limits, the final integral is much simpler than the first as: $$\frac{\rho_0}{4\pi\epsilon_0}\int^{\frac{1}{2}\pi}_0\int^{\frac{1}{2}\pi}_0\int^\infty_0\frac{\text{e}^{-r^2(1+\sin^2\phi)}}{r}r^2\sin\phi dr d\phi d\theta$$ Implementing all formula, we find that: $$\frac{\rho_0}{4\pi\epsilon_0}\int^{\frac{1}{2}\pi}_0\int^{\frac{1}{2}\pi}_0\int^\infty_0r\text{e}^{-r^2(1+\sin^2\phi)}\sin\phi dr d\phi d\theta$$ We can also allow $u=r^2$ to finally simplify the integral to: $$V_0=\frac{\rho_0}{8\pi\epsilon_0}\int^{\frac{1}{2}\pi}_0\int^{\frac{1}{2}\pi}_0\int^\infty_0\text{e}^{-u(1+\sin^2\phi)}\sin\phi du d\phi d\theta$$ Evaluating past the first integral, we find: $$V_0=\frac{\rho_0}{8\pi\epsilon_0}\int^{\frac{1}{2}\pi}_0\int^{\frac{1}{2}\pi}_0\frac{\sin\phi}{1+\sin^2\phi} d\phi d\theta=\frac{\rho_0}{16\epsilon_0}\int^{\frac{1}{2}\pi}_0\frac{\sin\phi}{1+\sin^2\phi} d\phi=\frac{\rho_0}{16\epsilon_0}\int^{\frac{1}{2}\pi}_0\frac{\sin\phi}{2-\cos^2\phi} d\phi$$ Using the subs $v=\cos\phi$, we finally obtain the single integral: $$V_0=\frac{\rho_0}{16\epsilon_0}\int_0^1\frac{1}{2-v^2} du$$ I trust you can finish off the integral from here! It was just a matter of finding the right parametrization for the specific charge density