Show that
$$-\frac 1 2 \int d^3x'\big (-i \phi(x', t)\nabla'^2\delta^3(x-x') \big ) = i\frac 1 2 \nabla^2\phi(x, t)$$
Attempt
$$-\frac 1 2 \int d^3x'\big (-i \phi( \vec x', t)\nabla'^2\delta^3(\vec x-\vec x') \big ) $$ $$=\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') -\frac{i}{2} \int d^3 x' \nabla' \phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x')$$ $$=\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x') + \frac{i}{2} \int d^3 x' \nabla'^2 \phi(\vec x', t) \delta^3(\vec x-\vec x')$$ $$= \frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x') - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')+\frac{i}{2}\nabla^2 \phi(\vec x, t)$$
I've been told that $\frac{i}{2}\phi(\vec x', t) \nabla' \delta^3(\vec x-\vec x'), - \frac{i}{2} \nabla'\phi(\vec x', t) \delta^3(\vec x-\vec x')$ terms vanish due to 'assuming boundary conditions'. However I do not see how to really show that such statement is true. How to do so?