Green function Helmholtz differential equation, monodimensional limit

1. Feb 6, 2013

galuoises

The solution of the problem
$\left(\nabla^2 + k^2 \right)\psi(\mathbf{r})=f(\mathbf{r})$ is, using green function

$\psi(\mathbf{r})=-\int G(\mathbf{r},\mathbf{r}_1) f(\mathbf{r})$
where for the tridimensional case the Green function is
$G(\mathbf{r},\mathbf{r}_1)=\frac{e^{ik\vert\mathbf{r}-\mathbf{r}_1\vert}}{4\pi \vert \mathbf{r}-\mathbf{r}_1\vert}$

Why if I put $f(\mathbf{r})=h(x)\delta(y-y_1)\delta(z-z_1)$ in the second equation I find
$G(\mathbf{r},\mathbf{r}_1)=\frac{e^{ik\vert x-x_1\vert}}{4\pi \vert \mathbf{x}-\mathbf{x}_1\vert}$
that it isn't the correct Green function for the monodimensional case, that is

$G(\mathbf{r},\mathbf{r}_1)=\frac{i}{2 k}e^{ik\vert x-x_1\vert}$