I am trying to solve the integral:
$$I = \int^{+\infty}_{-\infty} \mathrm{d} x \int^{+\infty}_{0} \mathrm{d} y \frac{f(x,y)}{g(x,y)}$$
where $g(x,y)$ has two zeros: $g(x=a,y)=0$ and $g(x,y=b)=0$, and both $a,b>0$. Using the Sokhotski-Plemelj formula:
$$\int \mathrm{d} z \frac{f(z)}{z} = - \mathrm{i} \pi \int \mathrm{d} z f(z) \delta(z) + \mathcal{P} \int \mathrm{d} z \frac{f(z)}{z}$$
For my problem, I want the resonant contribution from the integral:
$$\mathcal{Im} (I) = - \pi \int^{+\infty}_{-\infty} \mathrm{d} x \int^{+\infty}_{0} \mathrm{d} y f(x,y) \delta (g(x,y)) $$
Say $g(x,y) = c_1 - c_2 x(c_3+y)$. Therefore, $a = \frac{c_1}{c_2(c_3+y)}$ and $b = \frac{c_1}{c_2 x} - c_3$
I now try to rewrite the dirac delta in terms of each zero:
$$\mathcal{Im} (I) = - \pi \int^{+\infty}_{-\infty} \mathrm{d} x \int^{+\infty}_{0} \mathrm{d} y f(x,y) \frac{\delta (y-b) \delta(x-a)}{\sqrt{ \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2}}$$ $$ = - \frac{ \pi}{c_2} \int^{+\infty}_{-\infty} \mathrm{d} x \int^{+\infty}_{0} \mathrm{d} y f(x,y) \frac{\delta (y-b) \delta(x-a)}{\sqrt{ c_3^2 + 2 c_3 y + x^2 + y^2}} $$
Evaluating the integral over $y$ and $x$:
$$\mathcal{Im} (I) = - \frac{\pi}{c_2} \int^{+\infty}_{-\infty} \mathrm{d} x f(x,b) \frac{\delta(x-a)}{\sqrt{ c_3^2 + 2 c_3 b + x^2 + b^2}} = - \frac{\pi}{c_2} \frac{ f(a,b) }{\sqrt{ c_3^2 + 2 c_3 b + a^2 + b^2}}$$
Simplifying the denominator (in terms of $a$): $$\mathcal{Im} (I) = - \frac{\pi}{c_2} \frac{ f(a,b) c_2 a}{c_1} = - \pi \frac{ a}{c_1} f(a,b) $$
This answer doesn't seem right. I think my confusion comes from the fact $a$ and $b$ are functions of each other. I think I messed up the conversion of the dirac delta function into two seperate deltas. I think I should use a surface integral and parameterise, but I can't work out how. Does anyone know how to fix it?