Fourier transform of bubble momentum space integral I encountered the following integral in dimensional regularization
$$
I=\int d^d k \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{\vec{k}^2}\frac{1}{(\vec{q}-\vec{k})^2},
$$
say that we already Wick rotated the integral.
This looks like something which would be possible to evaluate but I'm not managing to do so. I tried Schwinger parametrization but got nowhere so far. I know it's possible to evaluate the integral at $x=0$ but I was wondering if anyone know a closed expression which is also a function of $x$. This is essentially the Fourier transform of the bubble diagram.
UPDATE
Here's where I got so far. Using a Feynman parameter $I$ can be written as
$$
I=\int {d^d k}  \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{\vec{k}^2}\frac{1}{(\vec{q}-\vec{k})^2}= \int_0^1  dt \int {d^d k} \,e^{i\vec{k}\cdot \vec{x}}\frac{1}{(\vec{k}^2-2t \vec{k}\cdot {q +t \vec{q}^2} )^2}\\=   \int_0^1  dt \, e^{it \vec{x}\cdot \vec{q}} \int d^d k e^{i\vec{k}\cdot \vec{x}} \frac{1}{(\vec{k}+ \Delta)^2}, \,\,\, \Delta=t(1-t)\vec{q}^2.
$$
Where I have shifted integration variable $k \to k-x q$ in the last step.
Using the suggested result  the loop integral should be (not being careful about numerical factors)
$$
\int d^d k\, e^{i\vec{k}\cdot \vec{x}} \frac{1}{(\vec{k}+ \Delta)^2}= |\vec{x}|^{2-d/2} (t(1-t)\vec{q}^2)^{d/4-1} K_{d/2-2}( \sqrt{t(1-t)} |\vec{q}||\vec{x}|),
$$
The last $t$ integral looks then pretty nasty then
$$\int_0^1 dt \, e^{it \vec{x}\cdot \vec{q}}   (t(1-t))^{d/4-1} K_{d/2-2}( \sqrt{t(1-t)}|\vec{q}||\vec{x}|), $$
maybe it can be done using residues and the integral representation on $K$? This looks weird since in the end I want to take $d=3$ but what's $K_{-1/2}$ then?
 A: The Fourier transform given by
$$
F[\Delta](\vec{x})=\int \mathrm{d}^dk\, e^{i\vec{k}\cdot\vec{x}} \frac1{(\vec{k}{}^2+\Delta)^2}\,,
$$
can be turned into the Hankel transform for the radial function $f(k) = 1/(k^2+\Delta)^2$ times a factor $k^{d/2-1}$. Namely
$$
F[\Delta](\vec{x})= \frac{(2\pi)^{d/2}}{x^{d/2-1}} \int_0^\infty k^{d/2-1}f(k)J_{d/2-1}(kx)\,k\,\mathrm{d}k\,,
$$
with $x \equiv (\vec{x}{}^2)^{1/2}$ and $J_\nu(z)$ is the Bessel function of the first kind.
In the linked Wikipedia page you can find a table with common results. If there is not what you need, some valus are also tabulated in [1].
In the aforementioned reference one finds that
$$
\int_0^\infty \frac{r^\nu}{(r^2+a^2)^{\mu+1}}\,J_\nu(xr)\,r\,\mathrm{d}r = \frac{a^{\nu-\mu} x^\mu K_{\nu-\mu}(as)}{2^\mu\Gamma(\nu+\tfrac12)}\,,
$$
where $K_\alpha$ is a modified Bessel function of the second kind and $\Gamma$ the Gamma function. Now just set $\mu=1,\, \nu=d/2-1,\,a^2 = \Delta$.

[1] A. D. Poularikas, Handbook of Formulas and Tables for Signal Processing. CRC Press, 1998.
A: Are you sure that you want that specific integral?  It looks like you are trying to compute something like
$$
\int d^d x e^{-ir(x-y)} [g(x,y)]^2 
$$
where
$$ 
g(x,y) = \int \frac{d^dk}{(2\pi)^d} \frac{ e^{ik(x-y)}}{k^2},
$$
is the massless propagator. Translation invariance makes this independent of $y$, and
gives
$$
 \int \frac{d^dk}{(2\pi)^d} \frac{1} {(k)^2(k-r)^2}= \Pi(r)
$$
without your exponential factor.  I can't think of any diagram computation that would give you the integral you are trying to do.
Perhaps what you are trying to do is compute
$$
\int \frac{d^dr}{(2\pi)^d} \int \frac{d^dk}{(2\pi)^d} e^{ir(x-y)} \frac{1} {(k)^2(k-r)^2}?
$$
This is
$$
[g(x,y)]^2= \int \frac{d^dr}{(2\pi)^d}e^{ir(x-y)} \Pi(r)
$$
and gives back the bubble in $x,y$  space.
As $g(x)\propto |x|^{-2(d-2)}$, you   can do any FT's of powers of $g$ by inverting
$$
 \int \frac{d^n k}{(2\pi)^n} e^{ik\cdot(x-x')} |k^2|^s = \frac{4^s}{\pi^{n/2}}\frac{\Gamma(s+n/2)}{\Gamma(-s)} \frac{1}{|x-x'|^{2s+n}}
$$
