Propagator proof

Question

Given $$ D(x-y)=\int\frac{d^4p}{(2\pi)^4}\ e^{ip(x-y)}\frac{i}{-p_\dot\ p - m^2} $$ How can I show that $$ (\partial^\mu\partial_\mu-m^2)D(x-y)=i\delta(x-y) $$ This looks like it must be trivial since my notes don't really give any proof of this.

Answer 1

Since a lot of people might want to check this really quick I will give my quick solution here.

We Fourier transform $(\partial^\mu\partial_\mu-m^2)$ which gives the value $$ \int\frac{d^4p}{(2\pi)^4} (-p^2-m^2)\ e^{ip(x-y)}\frac{i}{-p^2-m^2} = \int\frac{d^4p}{(2\pi)^4} i e^{ip(x-y)}=i\delta(x-y) $$