About an electrostatics integral and a delta-function kernel I'm having trouble with an integral and I would like some pointers on how to "take" it:
$$
\int \limits_{-\infty}^{\infty}\frac{3\gamma a^{2}d^{3}\mathbf r}{4 \pi \left( r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2} + a^{2}\right)^{\frac{5}{2}}}
$$
Here $\mathbf u$, $a$ and $\gamma$ are constants, and the integrand converges to a Dirac delta $\delta(\mathbf r)$ as $a\rightarrow 0$. The integral must be equal to 1.  
 A: First choose a direction for u, along the z-axis. Then the integral is
$$ I = \int {1\over (x^2 + y^2 + A z^2 + B)^{5/2} } dx dy dz $$
Rescale z by $\sqrt{A}$ to get rid of A and restore rotational invariance.
$$ I = {1\over \sqrt{A}} \int {1\over (x^2 + y^2 + z^2 + B)^{2.5}} dx dy dz $$
Now you do find the B dependence immediately from rescaling x y and z by $\sqrt{B}$ (or from dimensional analysis-- B has units length squared):
$$ I = {1\over \sqrt{A} B} \int {1\over (r^2 + 1)^{2.5}} d^3r $$
The only thing undetermined is the transcendental factor, which is just a number. You evaluate it by doing it radially and doing a string of substitutions:
$$ \sqrt{A}B I = 4\pi \int_0^{\infty} {r^2\over (r^2 +1)^{2.5} } dr $$
first $u=r^2 + 1$ gives
$$ \sqrt{A}B I = 2\pi \int_1^\infty {\sqrt{u-1}\over (u)^{2.5}} du $$
Then $v = {1\over u}$ makes it,
$$ \sqrt{A}B I = 2\pi \int_0^1 \sqrt{1-v} dv = {4\pi\over 3} $$
So
$$ I = {4\pi \over 3 \sqrt{A}B} $$
A: Set the $z$ axis along the direction of $\mathbf{u}$ and use spherical coordinates, which reduces your integral to something like
$$\int_0^\infty dr\int_0^\pi d\theta\int_0^{2\pi}d\phi\frac{r^2 \sin(\theta)}{\left(a^2 +r^2(1+\frac{\gamma^2}{c^2}\cos^2(\theta))\right)^{5/2}}.$$
Do the $\phi$ integral first and then the $\theta$ integral, transforming to $u=\cos(\theta)$. Be careful to use absolute values for the roots when necessary. After that the $r$ integral should be tough but doable.

EDIT to take some discussion off comments.
I gave 

Integrate[Sin[[Theta]]/(a^2+r^2 (1+[Gamma]^2 Cos[[Theta]]^2))^(5/2),[Theta]]/.[Theta]->[Pi]

to Mathematica to get
$$\frac{3 \left(a^2+r^2\right)+2 r^2 \gamma ^2}{3 \left(a^2+r^2\right)^2 \left(a^2+r^2+r^2 \gamma ^2\right)^{3/2}},$$
and then doing the radial integration by

Integrate[r^2 (3 (a^2+r^2)+2 r^2 [Gamma]^2)/(3 (a^2+r^2)^2 (a^2+r^2+r^2 [Gamma]^2)^(3/2)),r]

gives 
$$\frac{r^3}{3 a^2 \left(a^2+r^2\right) \sqrt{a^2+r^2 \left(1+\gamma ^2\right)}}$$
for the antiderivative. I agree that the roots make one suspect nonelementary antiderivatives but it is only one root so that elliptic integrals are out. Once one has the antiderivative, of course, it is routine to check that it does differentiate to what it should.
