# Fourier transform (2D boson propagator in Euclidean space)

Can anyone help me with the following Fourier transforms in 2D? $$\int \frac{e^{i\vec{k}\cdot\vec{r}}}{|\vec{k}|^2+M^2}~dk_x dk_y,$$ $$\int \frac{e^{i\vec{k}\cdot\vec{r}}}{(|\vec{k}|^2+M^2)^2}~dk_x dk_y$$ Or maybe some useful handbooks on multidimensional Fourier transforms?

• The first is $2\pi K_0(m r)$ with $K_0(x)$ the modified Bessel function. The second integral, I do not know from the top of my head. Commented Nov 16, 2016 at 16:44
• @Fabian Hi, thanks! Where could I find the derivation of the first one? Commented Nov 16, 2016 at 16:54

The first integral is $$2 \pi K_0(M r).$$ The second integral is $$\frac{\pi r}{M} K_1(M r),$$ where $K_m$ is the modified Bessel function of the second kind with $r=|\bf r|$. I will either prove a reference or outline the derivation shortly.
Here is a derivation. We start with the problem $$I_a = \int\!d^2k\,\frac{e^{i {\bf k}\cdot{\bf r}}}{(M^2 + |{\bf k}|^2)^a} = \int_0^\infty\!dk\, \frac{k}{(M^2 + k^2)^a} \int_0^{2\pi}\!d\phi \,e^{i k r \cos(\phi)}$$ with $k=|{\bf k}|$ and $r=|{\bf r}|$ and $\phi$ the angle between $\bf k$ and $\bf r$. The integral over $\phi$ yields $$\int_0^{2\pi}\!d\phi \,e^{i k r \cos(\phi)} =2 \pi J_0(kr) = 2\pi \mathop{\rm Re} H_0^{(1)}(k r).$$
We are left with the problem ($a=1,2$) $$I_a = \mathop{\rm Re}\int_0^\infty\!dk\,\frac{2 \pi k H_0^{(1)}(k r)}{(M^2+k^2)^a}= \int_{-\infty}^\infty\!dk\,\frac{\pi k H_0^{(1)}(k r)}{(M^2+k^2)^a}.$$ Now we employ the residue theorem for the contour along the real axis closed by a large semicircle in the positive imaginary plane. The integral along the semicircle vanishes (as $H_0^{(1)}$ goes to zero exponentially in the upper half-plane). We obtain $$I_a = 2 \pi i \mathop{\rm Res}_{k=i M}\frac{\pi k H_0^{(1)}(k r)}{(M^2+k^2)^a}.$$ In the case $a=1$, we have a simple pole with the result $$I_1 = 2 \pi i \frac{\pi i M H_0^{(1)}(i M r)}{2 i M }= 2\pi K_0(M r)$$ as $H^{(1)}_0(i x)= (2/i\pi) K_0(x)$.
For $a=2$, we have a double pole and after a short calculation we find $$I_2 = 2\pi i \frac{-i \pi \partial_xH_0^{(1)}(x=iMr)}{4M}= -\frac{\pi^2 r H_1^{(1)}(i M r)}{2M}= \frac{\pi r}{M} K_1(M r).$$
• Thanks. But one question:$I_a = \mathop{\rm Re}\int_0^\infty\!dk\,\frac{2 \pi k H_0^{(1)}(k r)}{(M^2+k^2)^a}= \int_{-\infty}^\infty\!dk\,\frac{\pi k H_0^{(1)}(k r)}{(M^2+k^2)^a}.$ Isn't the Hankel function $H_0^{(1)}(x)$ even so that the right-hand side always vanish? This is the problem that stucked me when I tried going this direction. Commented Nov 16, 2016 at 19:18
• The real part of $H^{(1)}_0$ is odd and the imaginary part even, see here. Commented Nov 16, 2016 at 19:50