I am calculating a commutator relation over all space and I obtain an integral:
$\int d^3p \nabla_{x - y} e^{ip\cdot (x - y)}$. Here $x,y,p$ are three vectors. I am wondering if the following is valid:
$\int d^3p \nabla_{x - y} e^{ip\cdot (x - y)} = \nabla_{x - y}\bigg[\int d^3p e^{ip\cdot (x - y)}\bigg] = \nabla_{x - y}\bigg[\int d^3p e^{-ip\cdot (y - x)}\bigg] = (2\pi)^3\nabla_{x - y} \delta^{(3)}(y - x) = (2\pi)^3\nabla_{x - y} \delta^{(3)}(x - y)$.
Here I used the fact that $\int d^3p e^{-ip(x - y)} = (2\pi)^3 \delta^{(3)}(x - y)$.
My Motivation: I was planning to use the volume integral of a gradient of $\phi$ is a surface integral of $\phi$ but I noticed that $e^{ip\cdot (x - y)}$ doesn't really have a value at the surface (i.e. spherical shell with radius $p = \infty$). To see this, one can use $e^{ix} = cosx + isin(x)$.
So my question is: Can I pull out the gradient from the integrand and have the gradient act on the entire integral? If not, how do I evaluate this integral?
Note: the gradient of a Dirac delta function is valid in physics.
First, the object $\int_{\mathbb{R}^3}\nabla_{\vec x-\vec y}e^{i\vec p\cdot (\vec x-\vec y)}\,d^3p$ is not an integral. Rather it represents a distribution that is equal to the Fourier Transform of $i\vec p$, written $\mathscr{F}\{i\vec p\}(\vec x-\vec y)$.
Now, it is true that
$$\begin{align} \mathscr{F}\{i\vec p\}(\vec x-\vec y)&=\nabla_{\vec x-\vec y}\mathscr{F}\{1\}(\vec x-\vec y)\\\\ &=(2\pi)^3 \nabla_{\vec x-\vec y} \delta^3(\vec x-\vec y)\tag1 \end{align}$$
So, if we use the more common and abusive notation of integrals, then we can write $(1)$ as
$$\int_{\mathbb{R^3}}(i\vec p)e^{i\vec p\cdot (\vec x-\vec y)}\,d^3p= \nabla_{\vec x-\vec y}\int_{-\infty}^\infty e^{i\vec p\cdot (\vec x-\vec y)}\,d^3p\tag2$$
whence which we conclude from $(2)$ that
$$\int_{\mathbb{R^3}}\nabla_{\vec x-\vec y}e^{i\vec p\cdot (\vec x-\vec y)}\,d^3p= \nabla_{\vec x-\vec y}\int_{-\infty}^\infty e^{i\vec p\cdot (\vec x-\vec y)}\,d^3p$$