Missing a $\frac{-2}{\pi}$ factor on the Green's function for the 2D Laplacian I want to prove that one of the possible Green's function for the 2D Laplace's equation in Cartesian coordinates, which solves:
$$(\partial_x^2+\partial_y^2)G=\delta(x-x_0)\delta(y-y_0)$$
With the boundary conditions
$$G(0,y)=G(x,0)=G(x,b)=0, \quad G(\infty,y)<\infty$$
By finding the solutions of the homogenous Laplace's equation$(\nabla^2y=0) $ with the above conditions I get: (I can elaborate on the process if needed)
$$
G= \begin{cases}
\int_0^\infty \frac{\sin\left(kx_0\right)\sin\left(kx\right)\sinh\left(ky\right)\sinh\left(k(b-y_0)\right)}{k\sinh(kb)}dk & y<y_0 \\
\int_0^\infty \frac{\sin\left(kx_0\right)\sin\left(kx\right)\sinh\left(ky_0\right)\sinh\left(k(b-y)\right)}{k\sinh(kb)}dk & y>y_0
\end{cases}
$$
The correct Green's function should be this exact same function, but multiplied by a factor of $\frac{-2}{\pi}$ at the beginning. I've been trying to figure out where this comes from, but to no avail. How do I get this evasive $\frac{-2}{\pi}$?
 A: I believe I have managed to solve it.
The problem was that the Green's function of the question doesn't have the right jump discontinuity on its derivative. For this equation, the jump has to be 1 (as the polynomial p(x) of the Sturm-Liouville form of the equation $\nabla(\nabla\cdot (p({\bf x})y({\bf x}))+q({\bf x})y({\bf x}$) is 1)
So:
$$
\partial_yG\vert_{y_0^+}-\partial_yG \vert_{y_0^-}=1
$$
Substituting G with an unknown constant A (which is what I want to find):
$$
1=\int_0^\infty \frac{A\sin(kx_0)\sin\left(kx\right)(-\sinh\left(ky\right)\cosh\left(k(b-y))\right)-\cosh(ky)\sinh(k(b-y_0)))}{\sinh(kb)}dk
$$
The two terms on y are just the hyperbolic sine of the sum of the arguments, which happens to be $\sinh(kb)$, and it cancels with the denominator. We end up with:
$$
1=-A\int_0^\infty sin(kx_0)sin\left(kx\right)dk
$$
Now I use the property $\sin(A)\sin(B)=\frac{1}{2}(\cos(A-B)-\cos(A+B)$ and apply the definition of the cosine as complex exponentials:
$$
\sin(x_0)\sin(x)=\frac{1}{2}(\cos(k(x-x_0))-\cos(k(x+x_0)))=\frac{1}{4}\left(e^{ik(x-x_0)}+e^{ik(x_0-x)}-e^{ik(x+x_0))}-e^{-ik(x+x_0))}\right)
$$
So the integral becomes:
$$
1=-A\int_0^\infty \frac{1}{4}\left(e^{ik(x-x_0)}+e^{ik(x_0-x)}-e^{ik(x+x_0))}-e^{-ik(x+x_0))}\right) dk
$$
Because the exponentials on the inside are just the cosines rewritten, this can be seen as one of the representations of the Dirac delta distribution (the inverse Fourier transform of 1) if we take $\int_0^\infty\cos x dx=\frac{1}{2}\int_{-\infty}^{\infty}\cos x dx$. I will take the convention for the transforms which divides by $2\pi$ when doing the inverse transform, so the exponentials alone are $2\pi$ times the delta.
$$
1=-A\frac{\pi}{4}(\delta(x-x_0)+\delta(x_0-x)-\delta(x+x_0)-\delta(-(x+x_0)))
$$
If we integrate this on an interval $[x_0-\epsilon,x_0+\epsilon]$ and take $\lim_{\epsilon\rightarrow 0}$, only the first two deltas are nonzero, and their integral is 1. So:
$$
1=-A\frac{\pi}{2} \Longrightarrow A=\frac{-2}{\pi}
$$
I'm not $100$% sure that what I've done is completely justified/right, but it solves my problem. Any feedback is appreciated.
