I am considering the integral
$$ I = \int_{-\infty}^\infty \text{d} x \: \frac{f(x)}{x}e^{-x^2}, $$
where $f(x)$ is an even function of $x$ with no singularities/poles. Due to the (anti)symmetry of the integral, I am pretty sure that the Cauchy principal value of this integral is zero. However, I would like to prove this via contour integration. As such, I consider a related integral:
$$ J = \oint_{C} \text{d}z \: \frac{f(z)}{z}e^{-z^2}$$
where for some closed contour $C$. However, I am having trouble choosing the contour. One cannot adopt a semicircle in the upper half-plane (enclosing the pole) as the integral over the portion of the semicircle for $\theta \in [\pi/2,\pi]$ will not vanish by Jordan's Lemma. I thought that one could adopt a rectangular contour in the upper half-plane (once again enclosing the pole), running from $-R$ to $R$ along the real axis (going below the pole at $z=0$), and then from $-R$ to $R$ along $z=+ia$ for some positive real number $a$. However, I come unstuck when trying to manipulate the integral along the upper part of the rectangle.
Is my choice of contour appropriate for finding the Cauchy principle value of $I$, or is there a more appropriate contour? Any help would be greatly appreciated.