I highly suspect that $$\int_{0}^{\infty} \cos(\cosh x) \cosh (\alpha x) \, \mathrm dx$$ converges for $0\le \alpha <1$
(If true, it obviously also converges for $-1 < a <0$.)
I can show that the integral converges for $\alpha=0$: $$\int_{0}^{\infty} \cos(\cosh x) \, \mathrm dx= \int_{1}^{\infty} \frac{\cos (u)}{\sqrt{u^{2}-1}} \, \mathrm du$$ which converges by Dirichlet's test
I can also show that the integral doesn't converge for $\alpha=1$:
$$\int_{0}^{\infty} \cos(\cosh x) \cosh(x) \, \mathrm dx = \int_{1}^{\infty} \frac{u \cos (u)}{\sqrt{u^{2}-1}} \, \mathrm du$$ which doesn't converge since $\frac{u \cos (u)}{\sqrt{u^{2}-1}} \sim \cos (u)$ for large values of $u$
For other values of $\alpha$ between $0$ and $1$, I'm not sure what to do. I don't know how to express $\cosh (\alpha x)$ in terms of $\cosh (x)$.