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?



1 Answer 1


Spherical harmonics decomposition totally worked for me. Someone suggested this method, but was not able to find a message from him...

Given expression and related can be presented as a $r^{-1}(Y_{1,-1}-Y_{1,1})$. Using the plane wave expansion formula for $e^{i\bf{k}r}$ factor, we can lift integrals by angular variables $\theta$, $\phi$. Taking the integral of remaining radial part will not cause any difficulties.

That was very significant for me. It helped to solve the main equation of elasticity for cubic medium with a weak anisotropy two years ago. Attempts of previous researches were unsuccessful and contained mistakes. Here is articles written by my colleague and me http://www.jetp.ac.ru/cgi-bin/dn/r_154_1211.pdf and Springer.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .