What's the integral of the laplacian of a delta function? There is little to find about this topic, but so far I have gathered:
$$\int_{-\infty}^{+\infty} f(r-r')\,\nabla_{r'}\delta(r')\,\mathrm{d^3r'} = \nabla_rf(r)$$
which can be obtained by taking us of:

*

*partial integration

*the gradient theorem $\left(\int\nabla_{r'}\delta(r')\,\mathrm{d^3{r'}} = \delta(r')\right)$

*as well as $\nabla_{r'}f = -\nabla_r f$

*and the fact that $f$ vanishes for $r \to \pm\infty\\[12pt]$
Now to step it up could you theoretically simplify:
$$\int_{-\infty}^{+\infty} f(r-r')\,\Delta_{r'}\delta(r')\,\mathrm{d^3r'} = \quad ?$$
 A: The Dirac Delta is not a function and the object $\int_{\mathbb{R}^3}f(\vec r-\vec r') \nabla_{\vec r'}\delta(\vec r')\,d^3r'$ is not an integral.  Rather, the Dirac Delta is a distribution and the integral symbol represents a linear functional.
The derivative $d'$ of a distribution $d$ is defined such that for any function $\phi\in C_C^\infty$ the functional $\langle d',\phi\rangle =-\langle d,\phi\rangle$. Therefore, we have for $f\in C_C^\infty$
$$\begin{align}
\langle \nabla_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=-\langle \delta,\nabla_{\vec r'}  \tau_{\vec r}Nf\rangle\\\
&=-\left.\left(\nabla_{\vec r'}\tau_{\vec r}Nf(\vec r')\right)\right|_{\vec r'=0}\\\\
&=-\left.\left(\nabla_{\vec r'}f(\vec r-\vec r')\right)\right|_{\vec r'=0}\\\\
&=\nabla_{\vec r} f(\vec r)
\end{align}$$
where $N$ is the negation operator that takes $f(\vec r')$ to $f(-\vec r')$ and $\tau_{\vec r}$ is the translation operator that takes $f(-\vec r')$ to $f(\vec r-\vec r')$.
Then, in an analogous fashion we can write
$$\begin{align}
\langle \nabla^2_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=\langle \delta,\nabla^2_{\vec r'}  \tau_{\vec r}Nf\rangle\\\
&=\left.\left(\nabla^2_{\vec r'}\tau_{\vec r}Nf(\vec r')\right)\right|_{\vec r'=0}\\\\
&=\left.\left(\nabla^2_{\vec r'}f(\vec r-\vec r')\right)\right|_{\vec r'=0}\\\\
&=\nabla^2_{\vec r} f(\vec r)
\end{align}$$

If one is unfamilar with the notation of distributions, then one can conduct the analysis using heuristical, formal arithmetic as follows.
$$\begin{align}\require{cancel}
\langle \nabla^2_{\vec r'} \delta, \tau_{\vec r}Nf\rangle&=\int_{\mathbb{R}^3}f(\vec r-\vec r')\nabla^2_{\vec r'}\delta(\vec r')\,d^3r'\\\\
&=\int_{\mathbb{R}^3}f(\vec r-\vec r')\nabla_{\vec r'}\cdot \nabla_{\vec r'}\delta(\vec r')\,d^3r'\\\\
&=\cancelto{0}{\int_{\mathbb{R}^3}\nabla_{\vec r'} \cdot \left(f(\vec r-\vec r') \nabla_{\vec r'}\delta(\vec r')\right)\,d^3r'}\\\\
&-\int_{\mathbb{R}^3}\nabla_{\vec r'}  f(\vec r-\vec r)\cdot \nabla_{\vec r'}\delta(\vec r')\,d^3r'\\\\
&=-\cancelto{0}{\int_{\mathbb{R}^3}\nabla_{\vec r'}\cdot \left(\nabla_{\vec r'}  f(\vec r-\vec r)\delta(\vec r')\right)\,d^3r'}\\\\
&+\int_{\mathbb{R}^3} \nabla^2_{\vec r'}f(\vec r-\vec r')\delta(\vec r')\,d^3r'\\\\
&=\nabla_{\vec r}^2f(\vec r)
\end{align}$$
as expected!
