I'm trying to reproduce the results of a paper, I'll insert here what is important and then explain what I cannot understand.
We have this equation for the Green function $\mathcal{G}$: $$\left( D_y \frac{\partial^2}{\partial y^2}-v_y\frac{\partial}{\partial y} -\delta(y)D_xk_x^2-ik_xv_x-s \right) \mathcal{G} (k_x,y,\overline{y};s)=\delta (y- \overline{y}),$$ subjected to the boundary conditions $\mathcal{G}(k_x, \pm\infty, \overline{y};s)=0$. This equation can be solved by a Fourier transform with respect to $y$ (in the "physicist form": $\mathcal{F} [f(y)](k_y)=\int_{-\infty}^{\infty} dye^{-ik_y y}f(y)$ and $\mathcal{F}^{-1}[f(k_y)](y)=\frac{1}{2\pi} \int_{-\infty}^{\infty}dk_ye^{ik_yy}$ ): $$\mathcal{G} (k_x,k_y,\overline{y};s)=- \frac{e^{-ik_y\overline{y}}}{D_yk_y^2 +ik_yv_y+ik_xv_x+s}-\frac{D_xk_x^2}{D_yk_y^2 +ik_yv_y+ik_xv_x+s} \mathcal{G}(k_x,0,\overline{y};s).$$
The last term has the presence of the Green function at $y=0$, which after some calculation it is possible to show that is given by $$\mathcal{G}(k_x,0,\overline{y};s)=-\frac{e^{-\frac{v_y}{2 D_y}\overline{y}} e^{-\sqrt{\frac{\beta}{D_y}}\vert\overline{y}\vert }}{2\sqrt{D_y \beta} +D_xk_x^2},$$ where $\beta=s+ik_xv_x+v_y^2/(4D_y)$.
I can't reproduce these "some calculations" that give the expression for $\mathcal{G}(k_x,0,\overline{y};s)$. Maybe I have to solve the first equation and then apply $y=0$, because when he says "$y=0$" implies that it should be done before the Fourier transform; but I don't know how to do it.
PS: $D_y$, $D_x$, $v_y$ and $v_x$ are real constants.