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.


1 Answer 1


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 $d=2$.


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