I'm trying to take the integral of non-radial function in spherical coordinates.
$$\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty} \frac{x^2}{r^3} e^{2\pi i \vec{r}\vec{p}}dx dy dz= \int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{cos^2(\phi)sin^3(\theta)}{1}r e^{2\pi i p r \cos(\gamma)} dp d\phi d\theta$$
I can't just change the orientation of z-axis to simplifier computing the $\cos(\gamma)$ into $\cos(\theta)$, because of my function depends on angles $\phi$ and $\theta$.
I have an idea with Dirac delta-function to use its filter property. Smth like this: $$\int\int \int F(r,\theta,\phi) e^{2pi r p \cos(\gamma)}dpd \phi d\theta \Rightarrow \int\int \Phi(\theta, \phi)\text{} \frac{\delta(\cos(\gamma))}{p} \Rightarrow \int \Phi(\cos(\gamma)=0) $$ I actually know that itsn't accurate from the side of math, but a trick with delta function should help to reduce numbers of the integrals. There is a problem regarding the representation of dirac delta: $$\delta(p) = \int_{-\infty}^{\infty} e^{-2\pi i r p} dr$$ So, my $r$ changes only from $0$ to $\infty$ and it will not represent $\delta$ :(
How can I built a $\delta$ from $\int_{0}^{\infty}$?
Is there any way to take integrals like mine?
Thanks.
