Problem over a definite integral, which surely needs contour integration

Assuming $a >0$, the equation $$\int_{0}^{\infty} \frac{1}{x^2 + a^2} \, \cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right) \, dx = \frac{\pi}{2a} \, \exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right)$$ holds if $b^{2} \ge c^{2}$.

To see why this is the case, consider the function $$f(z) = \frac{1}{a^{2}+z^{2}} \, \exp \left(iz \, \frac{z^{2}-b^{2}}{z^{2}-c^{2}} \right) = \frac{1}{a^{2}+z^{2}} \, \exp(iz) \exp \left(-iz \, \frac{b^2-c^2}{z^2-c^2}\right). $$

(Expressing $f(z)$ in that alternative way, which was suggested to me by Daniel Fischer, makes the analysis a bit easier.)

In the upper half-plane, both $|\exp(iz)|$ and $ \left| \exp\Bigl(-iz \, \frac{b^2-c^2}{z^2-c^2}\Bigr) \right|$ are bounded if $b^{2} \ge c^{2}$.

The latter is not particularly obvious. But by substituting $x+iy$ for $z$, one finds that the real part of $-iz \, \frac{b^2-c^2}{z^2-c^2} $ is $$-\frac{(b^{2}-c^{2})(c^{2}y+x^{2}y+y^{3})}{(x^2-y^2-c^{2})^{2}+4x^2y^{2}},$$ which is never positive if $b^{2}\ge c^{2}$ and $y >0$.

So if $b^{2} \ge c^{2}$, the magnitude of $\exp\Bigl(-iz \, \frac{b^2-c^2}{z^2-c^2}\Bigr)$, like the magnitude of $\exp(iz)$, never exceeds $1$ in the upper half-plane (which includes near the essential singularities at $z= \pm c$).

Therefore, if $b^{2} \ge c^{2}$, we can integrate around a closed semicircular contour in the upper half-plane indented at $z=\pm c$ and conclude after taking limits that $$ \begin{align} \text{PV}\int_{-\infty}^{\infty} \frac{1}{x^2 + a^2} \, \cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right) \, dx &= \text{Re} \, 2 \pi i \, \text{Res} \left[f(z), ia \right] \\ &= \frac{\pi}{a} \, \exp\left(-\frac{a(a^2 + b^2)}{a^2 + c^2}\right). \end{align}$$

But since $\cos\left(\frac{x(x^2 - b^2)}{x^2 - c^2}\right)$ is bounded along the real axis, we can drop the Cauchy principal value sign.

I don't know the value of the integral when $b^{2} < c^{2}$.

I am going to stick my neck out, right here and now, and say that the answer provided by your professor's colleague is wrong. I did a simple numerical check in Mathematica and found the numerical result disagreeing with the proposed analytical result for $a=3$, $b=1$, and $c=2$.

Beyond that, however, it looks like whoever supplied this result merely used a semicircular contour in the upper half plane and used the residue theorem to find the residue of the integrand at the pole $z=i a$ and multiply by $i 2 \pi$ and be done with it. As nice and simple as it looks, this is wrong because the obvious essential singularity at $z=c$ has been ignored altogether. Further, the argument of the cosine term - to be replaced by an exponential - changes sign between $b$ and $c$, before, and after. It makes a difference whether $b \gt c$ or otherwise.

It is not obvious that this integral is a candidate for the residue theorem or any other scheme that uses integration in the complex plane. You need to examine all of the singularities of the integrand and see whether it makes sense to attack the integral in this manner. Further, you should do numerical checks on your answers before declaring victory.