Integration by parts with delta function in Klein-Gordon field derivation

[TroyElliott]

TL;DR

I need help understanding a term that appears in the derivation of the time dependent conjugate momentum of the Klein-Gordon field

When deriving $\Pi(\vec{x},t)$ for the Klein-Gordon equation (i.e. plugging $\Pi(\vec{x},t)$ into the Heisenberg equation of motion for the scalar field Hamiltonian), we come across a term that is the following
$\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y)$

We are then told "using integration by parts we get the following"

$\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y) = -\int_{-\infty}^{\infty}d^{3}y \delta(\vec{x}-\vec{y})\nabla^{2}\phi(y).$

For this to be true, the the boundary term $\delta(\vec{x}-\vec{y})\nabla\phi(y)$ evaluated at $\infty$ and at $-\infty$, must be zero. Are we assuming that the gradient of the the field goes to zero at infinity? And what about the delta function term when it becomes $\delta(\vec{x}-\infty)$? This blows up at the the boundary at infinity. Any thoughts on why this boundary term should be zero is much appreciated!

Thanks!

 

[Dr.AbeNikIanEdL]

I guess the simplest answer is that this is just the definition of the derivative for a $\delta$ distribution. This definition is of course designed to be compatible with partial integration if instead of the $\delta$ there would be an actual function. It does not really make sense to plug in numbers in the argument of $\delta$ like you would have to do for the boundary terms. We still have to assume that $\nabla\Phi$ vanishes at infinity for the integral to make sense in the first place.