= Solution
For real $b\ne0$, 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 is
$$
\operatorname{PV}\int_{-\infty}^{\infty}\frac{e^{iu}}{u^2-b^2}\,du=i\pi\left(\frac{e^{ib}}{2b}-\frac{e^{-ib}}{2b}\right)=-\frac{\pi\sin b}{b}.
$$
The imaginary part is odd and the real part even. Taking the real part and halving gives
$$
\boxed{\operatorname{PV}\int_0^\infty\frac{\cos u}{u^2-b^2}\,du=-\frac{\pi\sin b}{2b}.}
$$
The right-hand side is even in $b$, completing the negative-$b$ case as well.
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