Electric field of finite sheet: Full analytical solution of integration? I am trying to work out the integral
$$
\operatorname{E}_{z}\left(x,y,z\right) =
\alpha\int\int\frac{z\,\mathrm{d}x'\,\mathrm{d}y'}
{\left[\left(x - x'\right)^{2} + \left(y -y'\right)^{2} + z^{2}\right]^{3/2}}
$$ with the limits $$-\frac{a}{2}\leq dx'\leq\frac{a}{2}\qquad-\frac{b}{2}\leq dy'\leq\frac{b}{2}$$ Mathematica does not return the solution in a reasonable time and I can't seem to find it.
This integral is for the $z$ component of the electric field of a homogeneously charged finite sheets in the $z=0$ plane. The sheets dimensions are $a \cdot b$. The integrals that need to be solved are detailed on the second page of this text. Note that The three integrals are for $E_x$, $E_y$ and $E_z$ (typo in the linked document).
I solved the $E_x$ and $E_y$ integrals in Mathematica and I would be grateful for some help or a pointer to get the analytical expression for $E_z$.
 A: This integral may be done analytically as far as I can see.  First, for the inner integral over $y'$, use a trig substitution $y'=y+\sqrt{(z^2+(x-x')^2} \tan{\theta}$ to transform the integral into
$$\alpha z \int_{-a/2}^{a/2} \frac{dx'}{z^2+(x-x')^2} \, \int_{-\arctan{((b/2)+y)/\sqrt{(z^2+(x-x')^2}}}^{\arctan{((b/2)-y)/\sqrt{(z^2+(x-x')^2}}} d\theta \frac{\cos{\theta}}{\sin^3{\theta}}$$
The inner integral has a simple antiderivative, and after some algebra, we are down to single integrals:
$$\frac{\alpha z}{2} \left (\frac{b}{2}+y\right)^2 \int_{-a/2}^{a/2} \frac{dx'}{\left[ (x-x')^2+\left (\frac{b}{2}+y\right)^2+z^2\right ]\left[(x-x')^2+z^2 \right ]} - \\\frac{\alpha z}{2} \left (\frac{b}{2}-y\right)^2 \int_{-a/2}^{a/2} \frac{dx'}{\left[ (x-x')^2+\left (\frac{b}{2}-y\right)^2+z^2\right ]\left[(x-x')^2+z^2 \right ]}$$
Using partial fraction decompositions, we may simplify the above expression drastically to get
$$\frac{\alpha z}{2} \int_{-a/2}^{a/2} dx' \left [\frac{1}{(x-x')^2+\left (\frac{b}{2}-y\right)^2+z^2}- \frac{1}{(x-x')^2+\left (\frac{b}{2}+y\right)^2+z^2} \right ]$$
These integrals are expressible in terms of arctan, and I assume you can take it from here.
