I would like to evaluate the following integral:
$I = \int_{x_1}^{x_2} \int_{y_1}^{y_2} \int_{z_1}^{z_2} \frac{x-x_0}{[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2]^{3/2}} dz dy dx$
where the fixed point $(x_0,y_0,z_0)$ is interior to $[x_1,x_2]\times[y_1,y_2]\times[z_1,z_2]$. The issue is that the integrand becomes $\frac{0}{0}$ when $(x,y,z) = (x_0,y_0,z_0)$.
I've performed the integration in $x$ assuming $y=y_0$ and $z=z_0$ and get
$I = \frac{1}{|x_1|} - \frac{1}{|x_2|}$
Doing the same thing for $x$ and $y$ also gives convergent answers, so I think the whole thing should be convergent.
I have a solution for when $(x_0,y_0,z_0)$ is exterior to the integration region. Can I just use the same formula?
Yes you can. Proving the fact that the interior point in this question does not produce any problems is easy. For $\epsilon>0$, consider a spherical volume $S$ of radius $\epsilon$ centered at $(x_0,y_0,z_0)$. In spherical coordinates, the integral of $f(x,y,z)=\frac{x-x_0}{[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2]^{3/2}}$ over $S$ is $$I {= \iiint_S \frac{x-x_0}{[(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2]^{3/2}} dz dy dx \\= \iiint_S \frac{\rho\sin\theta\cos\phi}{\rho^3} \rho^2\sin\theta d\theta d\phi d\rho \\= \iiint_S \sin^2\theta d\theta d\phi d\rho=0, } $$ which implies convergence. Since $f(x,y,z)$ has odd symmetry w.r.t. $x=x_0$, it is possible to partition the interval $(x_1,x_2)$ into two intervals $J_1$ and $J_2$, where $J_1=(x_0-\epsilon_0,x_0+\epsilon_0)$ for some $\epsilon_0>0$ and $x_0\notin J_2$. Now, the integral over $J_1$ vanishes due to odd symmetry of $f(x,y,z)$ and for the integral over $J_2$, $(x_0,y_0,z_0)$ is now an exterior point.