I need to prove that $u(x,y,z,t)= \mathcal{O}(t^{-1}) $ uniformly as $t \to\infty$, or $t \cdot u(x,y,z,t) =$ some bound, using Kirchhoff's formula given by $\renewcommand{\vec}[1]{\mathbf{#1}} u(\vec{x}, t_{0}) = \frac{1}{4 \pi c^{2} t_{0}} \iint_S \psi(\vec{x}) \,dS + \frac{\partial}{\partial t_{0}} \left[ \frac{1}{4 \pi c^{2} t_{0} }\iint_{S} \phi (\vec{x}) \,dS \right]$
For $M_1 = \max\limits_{B(0,R)} \lvert \psi(x) \rvert$ where $\psi, \phi$ vanishes outside of $B(0,R)$
multiplying by $t_0$ on both sides, for $\left| \frac{1}{4 \pi c^{2}} \iint_S \psi(\vec{x}) \,dS \right| \leq \frac{M_1 R^2}{c^2}$
But for the second term in Kirchhoff's formula I get a $t_0$ term in the denominator after going through a similar procedure of setting a max
$M_2 = \max\limits_{B(0,R)} \lvert \phi(x) \rvert$
and then differentiating, at which point I end up with
$\left| \frac{\partial}{\partial t_{0}} \left[ \frac{1}{4 \pi c^{2} t_{0} }\iint_{S} \phi (\vec{x}) \,dS \right] \right| \leq -\frac{M_2 R^2}{c^2 t_0}$
I'm not even sure what I'm doing is right. The second term should not have the $t_0$ in the denominator I believe. Any guidance is appreciated.