Green function Helmholtz differential equation, monodimensional limit

 P: 8 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 $G(\mathbf{r},\mathbf{r}_1)=\frac{i}{2 k}e^{ik\vert x-x_1\vert}$ ?Can you help me please?