Let $$\Delta(x, m^2)=(2\pi)^{-3}\int e^{ip \cdot x}\theta(p^0)\delta(p^2+m^2)d^4p.$$
Here $\theta$ is the step function at $0$.
I would like to show that this is the same as $$\Delta(x, m^2)=(2\pi)^{-3}\int \frac{e^{ip \cdot x}}{2\sqrt{\vec{p}^2+m^2}}d^3p.$$
However I cannot see how to proceed. Could anyone help me? These two appear in the Weinberg book and treated as the same, but I cannot prove it myself.
I assume that there is a misprint, argument of delta-function should be $p^2-m^2$.
Consider case $p^0>0$, so the integral becomes $$\int d^4p\delta(p^2-m^2)e^{ipx}=\int dp_0\int d^3p\delta(p_0^2-p^2-m^2)e^{ip_0t}e^{ipx},$$ where zeros of delta-function are $$p_0=\pm\sqrt{p^2+m^2},$$ and delta function becomes $$\delta(p_0^2-(p^2+m^2))=\frac{\delta(p_0-\sqrt{p^2+m^2})}{2\sqrt{p^2+m^2}}-\frac{\delta(p_0+\sqrt{p^2+m^2})}{2\sqrt{p^2+m^2}}.$$ Performing integration over $p_0$, we obtain $$\int \frac{d^3p}{2\omega_p}e^{ip\cdot x},\quad \omega_p=\sqrt{p^2+m^2}.$$
What about term with $p_0=-\sqrt{p^2+m^2}$? Agument of square root is stricktly positive, so we forget about it.
