I had a look at the following post in the physics stack exchange about the coulomb field
In Pablo's answer he has the integral $$\int_{\mathbb{R}^3 } \frac{1}{(2 \pi )^3 } \frac{e^{i k \cdot r } }{k^2+a^2} d^3k$$
Which after switching to polar coordinates it becomes $$\frac{1}{(2 \pi )^3 } \int_0^{\infty } \frac{k^2}{k^2+a^2} dk \int_0^{2 \pi } d \phi \int_{-\pi /2 }^{\pi /2} \cos \theta e^{ikr \sin \theta } d \theta $$
However I can't understand how this was derived, since if $k=(k_1,k_2,k_3)$ and $r=((r_1,r_2,r_3)$ spherical coordinates are $$k_1 =\rho \cos \theta \sin \phi $$ $$k_2=\rho \sin \theta \sin \phi $$ $$k_3=\rho \cos \phi $$ so $$e^{ik \cdot r}$$ would become $$e^{i ( \rho \cos \theta \sin \phi r_1+\rho \sin \theta \sin \phi r_2 +\rho \cos \phi r_3 )} $$ I can't see how this resembles the given result, can you explain how is this derived?
Let $f(\vec r)$ be given by that integral
$$f(\vec r)=\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\frac{e^{i\vec k\cdot\vec r}}{k^2+a^2}\,d^3\vec k$$
With a fixed $\vec r$, we apply a rotational transformation in $\vec k$-space (i.e., $\vec k\mapsto \vec k'$) to align $k'_z$ with $\vec r$.
Note that $k^2=k'^2$ and $\vec k\cdot \vec r=k'r\cos(\theta')$. Then, inasmuch as the Jacobian of a rotational transformation is equal to $1$, we assert that
$$\begin{align} f(\vec r)&=\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\frac{e^{i\vec k\cdot\vec r}}{k^2+a^2}\,d^3\vec k\\\\ &=\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\frac{e^{ik'r\cos(\theta')}}{k'^2+a^2}\,d^3\vec k'\\\\ &=\frac1{(2\pi)^3}\int_0^{2\pi}\int_0^\pi \int_0^\infty \int_{\mathbb{R^3}}\frac{e^{ik'r\cos(\theta')}}{k'^2+a^2}\,k'^2 \sin(\theta')\,dk'\,d\theta'\,d\phi' \end{align}$$
Can you finish now?