 Solution

For real $b\ne 0$, the integrand depends only on $b^2$, so first take $b>0$. Integrate $e^{iz}/(z^2-b^2)$ over the upper-half-plane semicircle, indenting above both real poles with small clockwise semicircles. The large arc tends to zero by Jordan lemma. Each indentation contributes $-i\pi$ times its residue, and there are no enclosed poles. Thus the Cauchy principal value on the whole real axis isThe imaginary part is odd and the real part even. Taking the real part and halving givesThe right-hand side is even in $b$, completing the negative-$b$ case as well.