I'm self studying complex analysis, and I've found the following problem which I can't solve.
Prove that $$\int_{-\infty}^{\infty} e^{-x^2}Im(e^{-2ix}p(x+i))dx = 0$$
Where p(z) is a polynomial of real coefficients. As a hint the books says to consider the function $f(z)=e^{-z^2}p(z)$
I tried doing the integral $$\int e^{-z^2}p(z) dz $$ with the half circle countour of radius $R$ and doing $R \to \infty$ so that the integral fro $-\infty$ to $\infty$ appears, but I haven't had any luck. Any help solving this as well as the intuition behind the solution will be greatly appreciated.