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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?

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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

$$F(k) = \frac{4 \pi}{k} \int_0^{\infty} dr \, e^{-a r} \sin{k r} = \frac{4 \pi}{a^2+k^2}$$

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  • $\begingroup$ Interesting, I'll have to check my math there. As "a" goes to INF what is the limit of this? $\endgroup$ – Jackson Hart Aug 12 '15 at 1:56
  • $\begingroup$ Wait a second, the limit I had was -pi/2 to pi/2. you have 0 to pi. why the difference? $\endgroup$ – Jackson Hart Aug 12 '15 at 1:58
  • $\begingroup$ @JacksonHart: the 0 to $\pi$ limits are used with the sine factor to define the spherical coordinate system you want to use. For your limits you would use a cosine rather than a sine. $\endgroup$ – Ron Gordon Aug 12 '15 at 2:36
  • $\begingroup$ Could you clarify the end of post? Is the final thing the answer? Or is just the radial component? It is hard to follow for me $\endgroup$ – Jackson Hart Aug 12 '15 at 2:42
  • $\begingroup$ @JacksonHart: final thing is the answer. I skipped the steps because I felt it was not the point of the question. $\endgroup$ – Ron Gordon Aug 12 '15 at 2:46

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