How to Show the Integral of a Spherically Symmetric Potential?

[Demon117]

Homework Statement

Show that for a spherically symmetric potential

$\int _{all space} V(\vec{r})exp(i\vec{k}\cdot\vec{r})d\tau = \frac{4\pi}{r}\int_{0}^{\infty} V(r) sin(\kappa r)dr$

The Attempt at a Solution

Given that the potential is spherically symmetric we have azimuthal symmetry and zenithal symmetry, so that the integral reduces to

$\int _{all space} V(\vec{r})exp(i\vec{k}\cdot\vec{r})d\tau = 4\pi \int _{0}^{\infty}V(r)r^{2}exp(i\vec{k}\cdot\vec{r})dr$

From here, I am not sure how to work with the exponential portion. I've thought that perhaps since this is spherically symmetric we can reduce the dot product into $\vec{k}\cdot\vec{r}=kr cos(\theta)$ or something of that nature, but I really don't see how this helps me. If you have any suggestions or references that would help, please let me know. Thanks in advance.

 

[Redbelly98]

Homework Statement

Show that for a spherically symmetric potential

$\int _{all space} V(\vec{r})exp(i\vec{k}\cdot\vec{r})d\tau = \frac{4\pi}{r}\int_{0}^{\infty} V(r) sin(\kappa r)dr$

That can't be right. r is a variable of integration, it can't appear outside the integral.

The Attempt at a Solution

Given that the potential is spherically symmetric we have azimuthal symmetry and zenithal symmetry, so that the integral reduces to

$\int _{all space} V(\vec{r})exp(i\vec{k}\cdot\vec{r})d\tau = 4\pi \int _{0}^{\infty}V(r)r^{2}exp(i\vec{k}\cdot\vec{r})dr$

But the $\exp{(i\vec{k}\cdot\vec{r})}$ term spoils that symmetry, doesn't it?

From here, I am not sure how to work with the exponential portion. I've thought that perhaps since this is spherically symmetric we can reduce the dot product into $\vec{k}\cdot\vec{r}=kr cos(\theta)$ or something of that nature, but I really don't see how this helps me. If you have any suggestions or references that would help, please let me know. Thanks in advance.

You might try choosing a coordinate system where k lies along the z-axis, and set up the integral that way.