# Closed contour integral from Fourier Transform?

I have this problem from a physics book that i'm trying to understand..

$$\psi(\mathbf{r})=\frac{g}{(2 \pi)^3}\int d^3 p \frac{e^{i\mathbf{p\cdot r}}}{\mathbf{p}^2+\mu ^2}$$

The next steps i don't really understand what happens here, but this is what the author writes

$$=\frac{g}{(2 \pi)^2} \int_0^\infty p^2 dp \int_{-1}^{1} dz \frac{e^{iprz}}{p^2+\mu ^2}$$

Then i think the author uses contour integration here

$$=\frac{g}{4 \pi r} \frac{1}{2 \pi i} \int_{-\infty}^{\infty} e^{ipr} [\frac{1}{p+i \mu}+\frac{1}{p-i \mu}]=\frac{g e^{-\mu r}}{4 \pi r}$$

I mainly don't understand the 1st step..

• That is a change of coordinates to spherical ones, with the following replacement $\cos{\theta}\rightarrow{z}$ afterwards. – Kiryl Pesotski Feb 7 '18 at 13:43

## 1 Answer

The author implicitly verifies that $\psi$ is a radially symmetric function, because if $R\in O(3)$ is a rotation matrix then $$\begin{split} \psi(R\mathbf r) & = C\int d^3 p\, \exp(i R\mathbf r \cdot \mathbf p) \frac{1}{(R\mathbf p)^2 + \mu^2} &= C\int d^3 p\, \exp(i \mathbf r \cdot R^T\mathbf p) \frac{1}{(\mathbf p)^2 + \mu^2} \\ &= \psi(\mathbf r), \end{split}$$ where in the last step we used the variable change $R^T\mathbf p \mapsto \mathbf p$, that leaves $d^3p$ invariant.

So we may fix $\mathbf r = r\mathbf e_z$. Computing in spherical coordinates $p, \theta, \phi$ and setting $z=p\cos \theta$ (which implies that $d^3p = p^2 dp dz d\phi$) yields the first step of formula of the text.