I am wanting to find the transform of this:
$$f(r) = \frac{e^{-\alpha r}}{r}$$
where $r$ is the radial coordinate. And then I would like to find $\lim_{\alpha \to \infty}$.
I have this:
$\int_0^\infty dr \int_{-\pi/2}^{\pi/2} d\theta \int_0^{2\pi} d\phi r \sin{(\theta)} \exp{(-\alpha r + ikr \cos{(\theta)})}$
but when I do the integration I get 0.
Can someone see where I went wrong?
to expand: $2\pi\int_0^\infty dr \int_{-\pi/2}^{\pi/2} d\theta r \sin{(\theta)} \exp{(-\alpha r + ikr \cos{(\theta)})}$
then I do u-substitution
$$u=\cos{\theta}$$ $$du=-sin{\theta}$$
But after u-sub, I get 0 for the integral. Perhaps I made a mistake earlier?
Recall that the FT in 3D is
$$F(\mathbf{k}) = \int_{\mathbb{R}^3} d^3 \mathbf{r} \, f(\mathbf{r}) e^{i \mathbf{k} \cdot \mathbf{r}} $$
Now, we have radial symmetry. Thus, $F$ is radially symmetric and we may write
$$\mathbf{k} \cdot \mathbf{r} = k r \cos{\theta} $$
So we have
$$F(k) = 2 \pi \int_0^{\infty} dr \, r \, e^{-a r} \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{i k r \cos{\theta}} $$
The inner integral is, as the OP notes, doable by subbing the sine, or just evaluating as follows:
$$\int_0^{\pi} d\theta \, \sin{\theta} \, e^{i k r \cos{\theta}} = \frac1{i k r} \left (e^{i k r} - e^{-i k r} \right ) = \frac{2 \sin{k r}}{k r}$$
Note that the integral over $\theta$ is not zero; the OP's mistake is likely in not changing the sign of the cosine in the limits.
So we have
$$F(k) = \frac{4 \pi}{k} \int_0^{\infty} dr \, e^{-a r} \sin{k r} = \frac{4 \pi}{a^2+k^2}$$