I'm working on a proof and I've run into a step that I'm confident is correct, but I'm having trouble justifying. It starts with an expression that defines a distribution $D$ that has the following form: \begin{align} \frac{\partial^2 D}{\partial t^2} - \frac{1}{r_1^{d_1-1}}\frac{\partial}{\partial r_1}\left[r_1^{d_1-1}\frac{\partial D}{\partial r_1}\right] - \frac{1}{r_2^{d_2-1}}\frac{\partial}{\partial r_2}\left[r_2^{d_2-1}\frac{\partial D}{\partial r_2}\right] + \alpha^2 D &= \delta(t)\frac{\delta(r_1)}{\Omega_{d_1} r_1^{d_1-1}} \frac{\delta(r_2)}{\Omega_{d_2} r_2^{d_2-1}},\tag{1}\label{start} \end{align} where $\Omega_d= 2\sqrt{\pi^d} / \Gamma(d/2)$, and $\delta(x)$ is the Dirac delta function.
What I want to do is integrate both sides over all $r_2$, i.e. apply the operator $\int_0^\infty\mathrm{d}r_2\, r_2^{d_2-1} \Omega_{d_2}$. Because this is a definition, and the right hand side converges just fine, the left hand side must also. I can show that $\int_0^\infty\mathrm{d}r_2\, r_2^{d_2-1} D$ converges independently. Where I run into trouble is I believe that any of the other terms on the left hand side diverge if integrated individually on the conic sections defined by $r_2^2 = t^2 - r_1^2$ when $t > r_1$. That is, there is a cancellation among the terms. Basically, $D$ generally contains a singularity on the cones that are integrable for certain values of $d_1$ and $d_2$, but that are made worse when they are hit with two derivatives.
Once I can interchange the integrals with the partial derivatives that are in different variables and use the divergence theorem to turn the third term into a surface integral, everything would be fine. The trouble is that my understanding is that such interchanges of limits are only allowed when the integral converges. So, I have two questions:
- Can I combine the steps of splitting the integral term by term and interchanging the derivatives into a single step? The expression converges before-hand, and afterward, but having that intermediate step where things diverge seems suspect.
- Can I ignore the singularities in $D$ when apply the divergence theorem on term 2?
Basically, the goal is an expression of the form: \begin{align} \frac{\partial^2 D'}{\partial t^2} - \frac{1}{r_1^{d_1-1}}\frac{\partial}{\partial r_1}\left[r_1^{d_1-1}\frac{\partial D'}{\partial r_1}\right] - \oint \mathrm{d}\Omega_2 \ r^{d-1}\frac{\partial D}{\partial r_2} + \alpha^2 D' &= \delta(t)\frac{\delta(r_1)}{\Omega_{d_1} r_1^{d_1-1}}, \tag{2}\label{end} \end{align} with $D'\equiv \int_0^\infty \mathrm{d}r_2\,\Omega_{d_2}r^{d_2-1} D$. I can already prove that \eqref{start} and \eqref{end} converge individually. It's just justifying the transition between the two that is causing headaches.
One piece of additional information: when $r_1 > t$ the integrals all converge and everything is fine. That would suggest I could analytically continue the result from that region, but $D$ is not continuous across $r_2^2 = t^2 - r_1^2$, so I'm not sure that works, either.