This interesting integral arises in the calculation of wave reverberation in air ducts, and I believe it evaluates to: $$ \int_{-\infty}^\infty \frac{e^{ik\sqrt{x^2+w^2}}}{\sqrt{x^2+w^2}}\cos(\alpha x)\,dx = \pi i H_0^{(1)}\left(w\sqrt{k^2-\alpha^2}\right) $$ I can show this equation is true in the special cases when $\alpha = 0$ (a standard result -- change variables $u^2 = 1 + x^2/w^2$) and when $k = 0$ (express as integral along imaginary axis, then deform contour to loop around branch cut at $[1, \infty)$), both using the well-known property for the Hankel function: $$ H_0^{(1)}(x) = \frac{2}{\pi i}\int_1^\infty \frac{e^{ixt}}{\sqrt{t^2-1}}\,dt $$ Based on these two special cases, I guessed the general result above, which seems to hold true for all parameter values that I tried.
But how to prove it? Every contour I try blows up in one way or another. Alternatively, if I could just show that the answer is a function of $\sqrt{k^2-\alpha^2}$, then my special cases would generalize.
The integral, after replacing $\cos\alpha x$ with $e^{i\alpha x}$ and substituting $x=w\sinh t$, is equal to $$I=\int_{-\infty}^{\infty}e^{iw(k\cosh t+\alpha\sinh t)}\,dt.$$ Now if $|\alpha|<k$ then $k=r\cosh t_0$ and $\alpha=r\sinh t_0$ with $r=\sqrt{k^2-\alpha^2}$ and some $t_0$, and $$I=\int_{-\infty}^{\infty}e^{iwr\cosh(t+t_0)}\,dt=\int_{-\infty}^{\infty}e^{iwr\cosh t}\,dt=i\pi H_0^{(1)}(wr)$$ as expected (other cases are treated the same way).