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The Green's function for the 2D Helmholtz equation satisfies the following equation:

$$(\nabla^2+k_0^2+\mathrm{i}\eta)\,{\mathsf{G}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_o)=\delta^{(2)}(\mathbf{r}-\mathbf{r}').$$

By Fourier transforming the Green's function and using the plane wave representation for the Dirac-delta function, it is fairly easy to show (using basic contour integration) that the 2D Green's function is given by

$${\mathsf{G}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_0)=\displaystyle\lim_{\eta\to0}\int\frac{\mathrm{d}^2\mathbf{k}}{(2\pi)^2}\frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}')}}{k_0^2+\mathrm{i}\eta-k^2}=\frac{1}{4\mathrm{i}}\operatorname{H}_0^{(1)}\left(k_0|\mathbf{r}-\mathbf{r}'|\right)$$

where $\operatorname{H}_0^{(1)}$ is the Hankel function of zeroth order and first kind.

However, this 2D Green's function diverges (logarithmically) at $\mathbf{r}=\mathbf{r}'$. Therefore, if we want it to be well-defined for $\mathbf{r}=\mathbf{r}'$, one can introduce a Gaussian cut-off function like so

$$\tilde{{\mathsf{G}}}_{2\mathrm{D}}(\mathbf{r}-\mathbf{r}',k_0)=\displaystyle\lim_{\eta\to0}\int\frac{\mathrm{d}^2\mathbf{k}}{(2\pi)^2}\frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot(\mathbf{r}-\mathbf{r}')}}{k_0^2+\mathrm{i}\eta-k^2}\mathrm{e}^{-\frac{a^2k^2}{2}}$$

where $a$ is some cut-off parameter.

Question: How do you evaluate this integral?

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    $\begingroup$ Schwinger's trick? $\endgroup$
    – Sunyam
    Commented May 14, 2019 at 10:38
  • $\begingroup$ The same basic contour integration should work here. $\endgroup$
    – Roger V.
    Commented Apr 29, 2020 at 6:14
  • $\begingroup$ How did you solve the integral in the first place? I doesn't seem to be that easy tbh.. $\endgroup$ Commented Sep 30, 2020 at 19:51

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The poles are still in the same place so what is stopping you using residue theorem?

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    $\begingroup$ What contour do you have in mind? The gaussian piece is poorly behaved in the complex plane, as it diverges for $k=ix$ and $x\to \pm \infty$. $\endgroup$
    – user196574
    Commented Nov 29, 2022 at 17:56

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