I've been doing some mathematical modelling in geophysics and the following integration came up. Are there any identities for the Dirac Delta that can be used to simplify this expression:
$$\iint \delta \left( t-\frac1c\left(\sqrt{(x_1-z_1)^2+z_2^2}-\sqrt{z_1^2+z_2^2}\right) \right)dz_1dz_2$$
Edit: To make matters more interesting/challenging (or worse) there are other factors in the integrend, i.e., the full expression is:
$$\iint \frac{\rho(z_1)}{\sqrt{(x_1-z_1)^2+z_2^2}\sqrt{z_1^2+z_2^2}}\delta \left( t-\frac1c\left(\sqrt{(x_1-z_1)^2+z_2^2}-\sqrt{z_1^2+z_2^2}\right) \right)dz_1dz_2$$
where $\rho$ is a some random function. So the question now boils down to: Will Fubini together with
$$\delta(f(x)) = \sum_{i}\frac{\delta(x - x_i)}{|f'(x_i)|}$$
(as pointed out in the answers below) suffice for a simplification still?
Judging from the setup of the problem, I'll assume $x_1, t > 0$ and only study that case.
Let $\lambda = \frac12x_1$ and $\mu = \frac12 ct$. Introduce a bunch of variables:
$$ \begin{cases} (x,y) = (z_1 - \lambda, z_2)\\ r_1 = \sqrt{(x+\lambda)^2 + y^2} = \sqrt{z_1^2+z_2^2}\\ r_2 = \sqrt{(x-\lambda)^2 + y^2} = \sqrt{(z_1-x_1)^2 + z_2^2}\\ (u,v) = \left(\frac{r_1 - r_2}{2},\frac{r_1 + r_2}{2}\right) \end{cases}$$ The actual integral you want to integral can be rewritten as
$$\mathcal{I} = \int_{[-\lambda,\infty)\times(-\infty,\infty)}\rho(x+\lambda)\delta\left(\frac2c(u+\mu)\right)\frac{dxdy}{r_1r_2}$$
Since the integrand is symmetric with respect to the $y$-axis, we can replace the integral by one over the upper $(x,y)$ plane: $$\int_{[-\lambda,\infty)\times(-\infty,\infty)}\cdots\quad\leftrightarrow\quad 2\int_{[-\lambda,\infty)\times [0,\infty)}\cdots$$ Change variables to $(u,v)$, the upper $(x,y)$ plane corresponds to the strip $[-\lambda,\lambda] \times [\lambda,\infty)$ in the $(u,v)$ plane. Notice $$\begin{align} 4uv &= (r_1-r_2)(r_1+r_2) = r_1^2 - r_2^2 = 4\lambda x\\ u^2 + v^2 &= \frac14\left((r_1-r_2)^2+(r_1+r_2)^2\right) = \frac12\left(r_1^2+r_2^2\right) = x^2+y^2+\lambda^2 \end{align} $$ A point $(x,y) \in [-\lambda,\infty) \times [0,\infty)$ is equivalent to corresponding $(u,v)$ belongs to following region: $$\Omega \stackrel{def}{=} \{ (u,v) \in [ -\lambda,\lambda ] \times [ \lambda, \infty ) : uv \ge -\lambda^2 \}$$
In terms of $(u,v)$, the area element has the form $$dx \wedge dy = dx \wedge \frac{dy^2}{2y} = \frac{1}{2\lambda y} d(\lambda x)\wedge d(x^2+y^2+\lambda^2) = \frac{1}{2\lambda y} d(uv)\wedge d(u^2+v^2)\\ = \frac{1}{\lambda y} (udv + vdu)\wedge(udu + vdv) = \frac{v^2-u^2}{\lambda y} du\wedge dv = \frac{r_1r_2}{\lambda y} du\wedge dv $$ Since $$y^2 = u^2 + v^2 - x^2 - \lambda^2 = u^2 + v^2 - \frac{u^2v^2}{\lambda^2} - \lambda^2 = \frac{1}{\lambda^2}(v^2 - \lambda^2)(\lambda^2-u^2)$$ We find $$\frac{dxdy}{r_1r_2} = \frac{dudv}{\lambda y} = \frac{dudv}{\sqrt{(v^2-\lambda^2)(\lambda^2-u^2)}} $$ and hence $$\mathcal{I} = 2\int_\Omega \rho(x+\lambda)\frac{\delta(\frac{2}{c}(u+\mu))dudv}{\sqrt{(v^2-\lambda)^2(\lambda^2-u^2)}}$$
Apply the formula mentioned in other answers, we have $$\delta\left(\frac2c(u+\mu)\right) = \frac{c}{2}\delta(u + \mu)$$
Integrate the delta function over $u$, the $u$ get fixed to $-\mu$.
When $\mu > \lambda$, the line $u = -\mu$ lies outside the strip $[-\lambda,\lambda] \times [\lambda,\infty)$ and $\mathcal{I}$ vanishes.
For $0 < \mu < \lambda$, the integral reduces to
$$\mathcal{I} = \frac{c}{\sqrt{\lambda^2-\mu^2}} \int_\lambda^{\lambda^2/\mu}\frac{\rho\left(\lambda - \frac{\mu v}{\lambda}\right) d v}{\sqrt{v^2-\lambda^2}} = \frac{c}{\sqrt{\lambda^2-\mu^2}}\int_0^{\lambda-\mu} \frac{\rho(z_1)dz_1}{\sqrt{(\lambda-z_1)^2-\mu^2}} $$ The rest of the integral depends on the actual form of $\rho(\cdot)$ and I'll leave that for you.