Convergence of integral involving Laplacian Suppose that $u \in C^{2}(\bar{\Omega}_{0})$ with $\Omega_{0}$ a bounded domain in $\mathbb{R}^{3}$ and $ r_{0}=(x_{0},y_{0},z_{0})$ be a fixed point in  $\Omega_{0}$, then for $r=(x,y,z)$, how we can show that the integral $$ \int_{\Omega_{0}} \frac{ \nabla^{2} u(r)}{|r-r_{0}|} dv $$ is convergent?
My try: Because $\Omega_{0}$ is a bounded domain and $u$ is a $C^{2}$ function, therefore Laplacian will be bounded, but then how we can treat the remaining integral?
 A: Let $R>0$ be such that $\Omega_0 ⊂ B_R$ (the ball of size $R$ and center $0$). Then
$$
\begin{align*}
∫_{\Omega_0} \frac{\Delta u(r)}{|r-r_0|}\, \mathrm{d}r &≤ \|\Delta u\|_{L^\infty}
∫_{|r|<R} \frac{1}{|r-r_0|}\, \mathrm{d}r ≤ \|\Delta u\|_{L^\infty}
∫_{|r|<R+|r_0|} \frac{1}{|r|}\, \mathrm{d}r
\\
&\leq 4\pi \,\|\Delta u\|_{L^\infty} ∫_0^{R+|r_0|} \frac{1}{s}\, s^2\mathrm{d}s
\\
&\leq 2\pi \,\|\Delta u\|_{L^\infty} \,(R+|r_0|)^2
\end{align*}
$$
where I did the  change of variable $r ↔  r-r_0$, and then the radial change of variable with $s = |r|$ and $\mathrm{d}s = 4\pi\,s^2\mathrm{d}s$ (the volume of a sphere) (and with the notation $\Delta$ for the Laplacian, since for me $\nabla^2 = \nabla\nabla$ is the Hessian).

As a remark, one can lower the regularity assumptions on $u$ by using Hölder's inequality since for any $p>3/2$ and $p' = p/(p-1)$
$$
\begin{align*}
∫_{\Omega_0} \frac{\Delta u(r)}{|r-r_0|}\, \mathrm{d}r &≤ \|\Delta u\|_{L^p}
\left(∫_{|r|<R} \frac{1}{|r-r_0|^{p'}}\, \mathrm{d}r\right)^{1/p'}
\\
&\leq \left(\frac{4\pi}{3-p'}\right)^{1/p'} \,\|\Delta u\|_{L^p} \,(R+|r_0|)^{2-3/p}
\end{align*}
$$
