In deriving the Lorentz invariance of the measure

$$\int\frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}},$$

the following identity is used

$$\delta\left(x^2-x_0^2\right)=\frac{1}{2|x|}\left[\delta(x-x_0)+\delta(x+x_0)\right].$$

This is proved as follows. First note that for any integrable function $f(x)$, it holds that

$$\int_{-a}^af(x)\mathrm{d}x = \int_0^a\left[f(x)+f(-x)\right]\mathrm{d}x.$$

Now, consider the change of variables $x\to x^2$ with $\mathrm{d}(x^2)=2x \mathrm{d}x$. We find that

$$\int_\infty^\infty f(x)\mathrm{d}x = \int_0^\infty \left[f\left((x^2)^{1/2}\right)+f\left(-(x^2)^{1/2}\right)\right]\frac{\mathrm{d}(x^2)}{2(x^2)^{1/2}}\\
= \int_0^\infty\left[f(| x|)+f(-| x|)\right]\frac{\mathrm{d}(x^2)}{2| x|}.$$

To prove the desired identity we now compute

$$\int_{-\infty}^\infty f(x)\delta\left(x^2-x_0^2\right)\mathrm{d}x = \int_0^\infty \left[f(| x|)+f(-| x|)\right]\delta\left(x^2-x_0^2\right)\frac{\mathrm{d}(x^2)}{2| x|}\\
= \frac{f(| x_0|) + f(-| x_0|)}{2| x_0|}.$$

We also find that

$$\int_{-\infty}^\infty f(x)\frac{\delta\left(x-x_0\right)+\delta\left(x+x_0\right)}{2| x_0|}\mathrm{d}x = \frac{f(| x_0|) + f(-| x_0|)}{2| x_0|}.$$

Because this holds for any $f$, the desired identity follows.