Problem with inconsistent Fourier transform and inverse transform I'm using the convention that the fourier transform of a function $f(\vec x$) is$$\frac{1}{(2\pi)^{3/2}}\int{d^3x\: e^{-i\vec k\cdot \vec x}f(\vec x)}$$
I'm now fourier transforming the function $$f(\vec x)=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}$$ with $r:= \sqrt{x^2+y^2+z^2} $. The result I get is
$$\frac{\sqrt{(2a)^3}}{\pi}\frac{1}{(1+(ka)^2)^2}, \:   k:=\sqrt{k_1^2+k_2^2+k_3^2} $$
which I think should be correct. Once I do the backwards fourier transform (which looks exactly as the fourier transform just with an integration over $\vec k$ and $e^{i\vec k\cdot \vec x}$ instead of the negative exponent), I'm running into trouble. When I first start the inverse transformation I switch into spherical coordinates and after a few steps I arrive at:
$$\frac{\sqrt{(2a)^3}}{\pi} \sqrt{\frac{2}{\pi}}\frac{1}{r} \frac{2}{a^4}\int _0^\infty{ dk \frac{\sin(kr) k}{(k-\frac{i}{a})^2(k+\frac{i}{a})^2}}$$
I now want to solve this using the residue theorem. Since the function is even I can simply put an $\frac{1}{2}$ in front and integrate from $-\infty$ to $+\infty$ instead. Then the 2-fold degenerate complex zeros of the denominator are of course $z_1 = \frac{i}{a}$ and $z_2=-\frac{i}{a}$, with the following residues
$$Res(z_{1/2})=\mp \frac{1}{4}i a r \cosh{r/a}$$
If I understand the way to do these kind of integrals correctly, then I have to close the integral in the complex plane making it a contour integral and depending on wheter I close the loop in the positive or negative complex plane I will have to use the residue of $z_1$ or $z_2$ respectively, since that's the one that gets enclosed by the contour. But if I do it like that I simply get some kind of $\cosh$ for my inverse fourier transform and not the $e^{-r/a}$ that I started with.
I assume I'm just missing something about how to correctly solve these kind of complex integrals, since if I could somehow combine the two residues I could get the exponential I want out of them.
Any help or tipps are greatly appreciated, thank you!
 A: As mentioned in my comment, the integral over the part of the contour where $|z|\to\infty$ is not $0$. If you want to integrate over the upper contour, use $e^{ikr}$ instead of $\sin kr$. Then, in the upper half of the plane, $k$ becomes $|k|e^{i\phi}$. Then $$|e^{ikr}|=|e^{i|k|r(\cos\phi+i\sin\phi)}|=|e^{i|k|r(\cos\phi}e^{-|k|r\sin\phi}|$$
When $|k|\to \infty$ the second exponential goes to $0$. The absolute value of the first exponential is always $1$.
Then notice $$\int_{-\infty}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk=\int_{-\infty}^0\frac{ke^{ikr}}{(1+(ka)^2)^2}dk+\int_{0}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk$$
In the first integral replace $k$ with $-k$. Can you continue from here?
EDIT
$$\int_{-\infty}^\infty\frac{ke^{ikr}}{(1+(ka)^2)^2}dk=\int_0^\infty\frac{k(e^{ikr}-e^{-ikr})}{(1+(ka)^2)^2}dk=2i\int_0^\infty\frac{k}{(1+(ka)^2)^2}\frac{(e^{ikr}-e^{-ikr})}{2i}dk$$
The last fraction is $\sin(kr)$. Then using residue theorem, and the fact that on the upper contour the integral is zero, we get:
$$2i\int_0^\infty\frac{k\sin(kr)}{(1+(ka)^2)^2}=2\pi i \frac 14 ar e^{-r/a}$$
Notice that you have $2i$ on both sides. So when multiply with your prefactors you get $$\frac{\sqrt{(2a)^3}}{\pi} \sqrt{\frac{2}{\pi}}\frac{1}{r} \frac{2}{a^4}\frac 14\pi are^{-r/a}=\frac2{\sqrt{\pi a^3}}e^{-r/a}$$
This is your original function, up to a factor of $2$, that either you or I have missed in the calculation.
