How to derive the 2D asymmetric form of free Green function at large distance? The free Green function:
$$G(k)=\frac{1}{k^2+r}$$
I want to derive its form in real space(assuming the most trivial Euclidean metric) :
$$G(x)=\int d^D k \frac{e^{-ik\cdot x}}{k^2+r}$$
for two-dimension, i.e. $D=2$, it can simplified as:
$$G(x)\sim\int_0^\pi d\theta \int_0^\infty d k \frac{ke^{-ik x cos\theta}}{k^2+r}$$
I know this integral may not trivial, but I just want to derive its asymmetric  form  at large distance, i.e. $|x|\gg 1/\sqrt r$:
$$G(x)\sim( \sqrt{r}|\mathbf{x}|)^{-\frac{1}{2}} e^{-\sqrt{r}|\mathbf{x}|}$$
But I don't know the trick of integral to obtain it without using special functions.
 A: You can start from Schwinger parametrization,
$$\frac{1}{k^2+r}=\frac{1}{2}\int_0^{\infty}d\alpha\,\exp\left(-\frac{\alpha(k^2+r)}{2}\right),$$
which gives you
$$G(x)=\int_0^{\infty}d\alpha\int_k\exp\left[-\frac{\alpha(k^2+r)}{2}+ik\cdot x\right],\quad\int_k=\int\frac{d^dk}{(2\pi)^d}$$
and then just complete square, perform simple Gaussian integration and find
$$G(x)=\frac{1}{2(2\pi)^{d/2}}\int_0^{\infty}d\alpha\,\alpha^{-d/2}\exp\left(-\frac{x^2}{2\alpha}-\frac{r\alpha}{2}\right).$$
Setting $d=2$, we have
$$\boxed{G(x)=\frac{1}{4\pi}\int_{0}^{\infty}d\alpha\,\alpha^{-1}\exp\left(-\frac{x^2}{2\alpha}-\frac{r\alpha}{2}\right)}$$
In this integral we have large parameter $x$. Let us apply method of steepest descent. We rewrite integral as
$$G(x)=\frac{1}{4\pi}\int_0^{\infty}d\alpha\,\exp\left[-\ln\alpha-\frac{x^2}{2\alpha}-\frac{r\alpha}{2}\right]\equiv\frac{1}{4\pi}\int_0^{\infty}d\alpha\,\exp(f(\alpha)).$$
We state that integral saturates near the point $f'(\alpha_0)=0$. In the limit of large $x$, this point is
$$\alpha_0=x/\sqrt{r}.$$
Then,
$$f(\alpha_0)=-\sqrt{r}x-\ln(x/\sqrt{r}),\quad |f''(\alpha_0)|=\frac{r^{3/2}}{x}.$$
Approximate result for the integral is
$$G(x)\approx\frac{1}{4\pi}\sqrt{\frac{2\pi}{|f''(\alpha_0)|}}e^{f(\alpha_0)},$$
or explicitly
$$\boxed{G(x)\approx\frac{1}{2\sqrt{2\pi}}\frac{\sqrt{x}}{{\sqrt{r^{3/2}}}}\cdot\frac{\sqrt{r}}{x}e^{-\sqrt{r}x}=\frac{(\sqrt{r}x)^{-1/2}}{2\sqrt{2\pi}}e^{-\sqrt{r}x}}.$$
The exact answer for the integral is
$$G(x)=\frac{K_0(\sqrt{r}x)}{2\pi},$$
which has the same expansion for $x\rightarrow\infty$.
Hope it helps. Note that you can just set $d=2$ in the first line, but for me it was more comfortable to use arbitrary $d$ and then set $d=2$.
