I want to ask for assistance in verifying the fundamental solution of the wave equation in $\mathbb{R}^{3}$. Here the fundamental solution is given by $$\frac{1}{2\pi}H(t)\delta(t^{2}-|x|^{2})$$which according to the author should be equal to \begin{equation} \frac{1}{4\pi t}H(t)\delta(t-|x|) \end{equation}
I do not really understand why we had this change with extra $\frac{1}{2t}$ factor. I think since $t\ge 0$ because of the Heaviside function, we should have $\delta(t^{2}-|x|^{2})=2\delta(t-|x|)$instead. But still the $\frac{1}{t}$ term is missing.
To verify the second equation is a fundamental solution for the wave equation I have to check
$$\int_{0}^{\infty}\int_{\mathbb{R}^{3}}\frac{1}{4\pi t}H(t)\delta(t-|x|)(\partial_{t}^{2}-\Delta)F(t,x)dtdx=F(0,0)$$
Here are some problems I have. First it seems we are integrating over a 3D hypersurface instead of a 4D region. So by change of variables let $t=|x|=r$ the integrand $dtdx=r^{2}\sin[\theta]d\psi d\theta dr$. Second I am wondering if the radial transformation is the best way to evaluate the integral. The computation is really messy and does not seem to generalize to $n=20$ nicely. Here is the computation:
\begin{align} \frac{1}{4\pi}\int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0}\frac{1}{r}(\partial_{r}^{2}-\Delta)F[r,\theta,\psi]r^{2}\sin[\theta]d\theta d\psi dr \end{align} The middle part equals $$\partial_{r}^{2}-\Delta=\partial_{r}^{2}-\frac{1}{r^{2}}\partial_{r}(r^{2}\partial_{r})-\frac{1}{r^{2}\sin[\theta]}\partial_{\theta}(\sin(\theta) \partial_{\theta})-\frac{1}{r^{2}\sin^{2}[\theta]}\partial^{2}_{\psi}$$ After cancelling and multiplying $r^{2}\sin[\theta]$ from the Jacobian term we have $$-2r\sin[\theta]\partial_{r},-\partial_{\theta}(\sin(\theta) \partial_{\theta}),-\frac{1}{\sin[\theta]}\partial_{\psi}^{2}$$ So we have to evaluate the 3 integral respectively. The first one equals $$\frac{1}{4\pi}\int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0}\frac{1}{r}[-2r\sin[\theta]\partial_{r}F]d\theta d\psi dr$$ which equals $$=-\int^{\infty}_{0}\int^{\pi}_{0}\sin[\theta]\partial_{r}Fdr d\theta=-\int^{\pi}_{0}\sin[\theta]d\theta\int^{\infty}_{0}\partial_{r}Fdr=F[0,0]\int^{\pi}_{0}\sin[\theta]d\theta=2F[0,0]$$ the second integral equals $$\frac{1}{4\pi}\int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0}\frac{1}{r}-\partial_{\theta}(\sin(\theta) \partial_{\theta}F) d\theta d\psi dr=-\frac{1}{2}\int_{0}^{\infty}\frac{1}{r}dr(\sin(\theta) \partial_{\theta}F)|^{\pi}_{0}=0$$ since $\sin[\pi]=\sin[0]=0$.
the third integral equals $$\frac{1}{4\pi}\int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0}\frac{1}{r}-\frac{1}{\sin[\theta]}\partial_{\psi}^{2}F d\theta d\psi dr$$ where the evaluation of the innermost integral give us $0$ because $\partial_{\psi}F(2\pi)=\partial_{\psi}F(0)$.
Add up we have $2F[0,0]$ instead of $F[0,0]$. I am wondering which step I did wrong at here. I checked several times but did not spot a mistake.