Verifying the 2-dimensional fundamental solution of the wave equation I'm trying to verify that $$u(t,x)=H(t-|x|)(t^2-|x|^2)^{-1/2}$$ is the fundamental solution of the 2-dimensional wave equation; that is, $\Box u = u_{tt}-\Delta u = \delta_{0}$. I know there are tricks that make this calculation easier (involving realizing distributions as limits) but I'd really like to figure out how to do this with straightforward integration techniques.
I need to show that $$\langle u(t,x),\Box\varphi(t,x)\rangle = \langle\delta_0,\varphi(t,x)\rangle = \varphi(0,0,0),$$ where $\varphi\in C_{0}^{\infty}(\mathbb{R}^{1+2})$. So far I have:
$$\langle u(t,x),\Box\varphi(t,x)\rangle =\int_{-\infty}^{\infty}\int_{\mathbb{R}^2}H(t-|x|)(t^2-|x|^2)^{-1/2}\Box\varphi(t,x)dAdt$$
$$=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}H(t-r)(t^2-r^2)^{-1/2}r\Box_{\text{pol}}\varphi(t,r,\theta)drd\theta dt$$
$$=\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{t}\partial_{r}\left((t^2-r^2)^{\frac{1}{2}}\right)\left( \varphi_{tt} - \Delta_{\text{pol}}\varphi\right)drd\theta dt$$
$$=\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{t}\partial_{r}\left((t^2-r^2)^{\frac{1}{2}}\right)\left( \varphi_{tt}-\frac{1}{r}\partial_{r}\left(r\varphi_r\right)-\frac{1}{r^2}\varphi_{\theta\theta} \right)drd\theta dt.$$
Can anyone point me in the right direction? I'm sure integration by parts will be important here, but I'm not sure how to proceed. Any help?
 A: Fixed a small mistake...
I think the usual way is to express the fourier transfom of $u(x, t)$ as the (weak) limit of some "normal" function.  This is how the $i \epsilon$ appears in the physics books.  So what you should show is that
$$
u(x, t) = \mathcal{S}'-\lim_{\epsilon \rightarrow 0} \int \frac{dk^2 d\omega}{(2 \pi)^3} \frac{e^{-i(wt - (k, x))}}{|k|^2 - \omega^2 \pm i \epsilon}\, .
$$
Note that the sign of the $i \epsilon$ term is important.  Different ways of moving the poles away from the real axis give different fundamental solutions (moving forward and backward in time).  Follands book about fourier analysis shows how actual calculations are done.
A: I think the trick is to integrate by parts. 
We rewrite $H(t)(t^{2}-|x|^{2})^{-1/2}$ to be $$\lim_{\epsilon\rightarrow 0+}Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}$$
Then we need to verify that
$$F(0,0)=\lim_{\epsilon\rightarrow 0+}\int_{\mathbb{R}^{2}}\int^{\infty}_{0}Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}(\partial_{t}^{2}-\Delta)F(t,x)dtdx$$
If we mark $Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=u$, $(\partial_{t}^{2}-\Delta)F(t,x)=dv$. Then the $uv$ boundary part vanish because when $t=0$, $u=0$, and other boundary parts disappears. So we need to evaluate 
$$\lim_{\epsilon\rightarrow 0+}\int_{\mathbb{R}^{2}}\int^{\infty}_{0}Im(\partial_{t}+\partial_{x})(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}(\partial_{t}-\partial_{x})F(t,x)dtdx$$
We integrate by parts again. If we let $$Im(\partial_{t}+\partial_{x})(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=u', (\partial_{t}-\partial_{x})F(t,x)=dv$$
Then the $du'v$ term involves 
$$(\partial_{t}^{2}-\Delta)(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=0$$
So the only term really matters is the boundary term $u'v$. The $\partial_{x}$ part vanishes on the boundary, and what we are really integrating is
$$\int_{\mathbb{R}^{2}}\frac{\partial}{\partial t}(Im (|x|^{2}-(t+i\epsilon)^{2})^{-1/2})|_{t=0}F(0,x)dx$$
But this is the Possion integral on the upper half plane:
$$\int_{\mathbb{R}^{2}}\frac{\epsilon}{(|x|^{2}+\epsilon^{2})^{3/2}}F(0,x)dx$$
which tends to $F(0,0)$ times a constant when $\epsilon\rightarrow 0$. 
