I'm trying to understand the first step of the fourier transformation of the Yukawa potential.
The Yukawa potential is a spherically symmetric potential in 3 dimensions defined as $V_Y(r) = e^{-\alpha*r}/r$ with $r=|\vec{r}|$ and $\alpha >0$
We obtain the Fourier transformation by using polar coordinates (the z-axis points along the direction of k⃗):
\begin{align*} F(V_Y(r))&=\int_{\mathbb{R}^n}^{}\frac{e^{-\alpha*r}}{r}e^{-ik\vec{r}}d^3r \\ &=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\frac{e^{-\alpha*r}}{r}e^{-ikr\cos(\theta)}r^2\sin(\theta)dr d\theta d\phi \end{align*}
The rest is just super painful integration.
I have a very basic understanding of how Fourier works and what it's expressing but I don't have too much of an idea how the Fourier space looks like. In this first step, I don't get how the integral becomes a spherical integral but at the same time we use polar coordinates to express $e^{-ik\vec{r}}$ as $e^{-ikr\cos(\theta)}$ and how it holds?
Thank you in advance
"The z axis points along the direction of k"
so k has a direction and is a vector... $\vec k$, and the angle between it and $\vec r$ is the polar angle $\theta$. The exponent will be $$ exp(-i\vec k \cdot\vec r) $$ i.e a scalar product which is naturally $k \,r \cos \theta$.