I stumbled upon a tricky Fourier Transform that I got stuck on:
$$ \textbf I(\textbf{x}) = \int \frac{d^dp}{(2\pi)^d} ~e^{ - i \textbf x \cdot \textbf p} \frac{-i \textbf p}{p^2}\delta^d(\textbf p),$$
where $\delta^D$ is a Dirac delta in $d-$dim. Is the result simply $0$ for all $\textbf x$ or is there more to it?
P.S. I tried adding a $i \epsilon^2$ to the denominator (that gives zero in the limit $\epsilon \to 0$), but then the result changes if I add also something to the numerator... I'm confused.
I will apply the standard techniques of quantum field theory. Begin with, $$ I^\mu(x^\mu) = \frac{\partial}{\partial x_\mu} \int \frac{d^dp}{(2\pi)^d} \frac{e^{-ix\cdot p} }{p^2} \delta^{(d)}(p). $$ (I denote a vector with an index notation basically because I am used to that from physics.) Introduce a so-called Schwinger parameterisation, $$ I^\mu(x^\mu) = \frac{\partial}{\partial x_\mu} \int \frac{d^dp}{(2\pi)^d} \int_0^\infty d\tau e^{-ix\cdot p - \tau p^2} \delta^{(d)}(p). $$ Interchange(!) the order of integration which allows one to use the $\delta$-function, $$ I^\mu(x^\mu) = \frac{1}{(2\pi)^d}\frac{\partial}{\partial x_\mu}\int_0^\infty d\tau=0. $$ As is common with this technique, the integral is regularised to zero. I have not checked in detail here but generally speaking: It is permitted to interchange the order of integrals by completing the square of the exponential and moving to spherical polar coordinates then analytically continuing $d$ into the complex plane.