integration - Evaluate $int_{-infty }^{infty } left(cos left(sqrt{x^2-1}right)-cos left(sqrt{x^2+1}right)right) , dx$

From Gradshteyn&Ryzhik 3.692.6 we know that $$\int_{-\infty }^{\infty } \left(\cos \left(\sqrt{x^2-1}\right)-\cos \left(\sqrt{x^2+1}\right)\right) \ dx=\pi (J_1(1)+I_1(1))$$ Where two special functions on RHS is the original/modified Bessel function of the first kind (see here and here for their definitions if needed). I've scanned basic integral representations of these two functions, but none seems to have direct relationship with this problem. Also, according to the oscillatory behavior at infinity (in fact the plot of the integrand looks like a star), it can't be verified numerically by Mathematica.
I'd like you to help me and all kinds of methods are welcomed. Thank you.

Answer

Here's a sketch of a proof. I intend to use Ramanujan's Master Theorem (RMT), which states that for a function $F(x)=\sum_{k=0}^\infty \phi(k) (-x)^k/k!,$ with $\phi(0) \neq 0$, then

$$ \int_0^\infty x^{n-1} F(x) dx = \Gamma(n) \phi(-n). $$
This is true for where the integral converges.
A proof for non-integer n appears in L. Bougoffa, ArXiv 1902.01539v1, 5 Feb 2019. By splitting the integral at $x=0$ and scaling, we'll show the equivalent

$$ (1)\quad \int_0^\infty \Big( \cos{(\sqrt{2u-a^2}\ )} - \cos{(\sqrt{2u+a^2}\ )} \Big) \frac{du}{\sqrt{u}} =
\frac{\pi \ a}{\sqrt{2} }\Big( J_1(a) + I_1(a) \Big) .$$
The formula of the OP is the special case of $a \to 1.$ Naturally we will let $n=1/2$ in RMT so we need the Taylor expansion of the function in the big parentheses. I worked out the first 40 terms (with a symbolic computer program) and discovered a pattern:

$$ (2) \quad \cos{(\sqrt{2u-a^2}\ )} - \cos{(\sqrt{2u+a^2}\ )} =\sqrt{\pi a/2} \sum_{k=0}^\infty \Big(-J_{k-1/2}(a)+I_{k-1/2}(a)\Big) (-u/a)^k/k! $$
which means
$$ \phi(k) = \sqrt{\frac{\pi a}{2}} a^{-k} \Big(-J_{k-1/2}(a)+I_{k-1/2}(a)\Big). $$
Putting in $k=-1/2, \ \Gamma(1/2)=\sqrt{\pi}, $ and $ -J_{-1}(a) = J_1(a)$ gives the answer (1).

Of course I haven't proved (2), but it would surprise me if this expansion is not known. I have checked (1) numerically for many $0, using the PrincipalValue-> True argument in the numerical integration, with the point about which the principal value is taken at $u=a.$ Thus I think the suitable generalization is

$$
(3)\quad \int_{-\infty}^\infty \Big( \cos{(\sqrt{x^2-a^2}\ )} - \cos{(\sqrt{x^2+a^2}\ )} \Big) dx =
\pi \ a\Big( J_1(a) + I_1(a) \Big), \quad 0
as long as the integral is interpreted as a principal value.