I'm trying to evaluate the following integral
$$ \int_0^\infty dx ~\cos\left(\sqrt{1+x^2}a\right)~~~~(1) $$ where $a>0$. I've noticed that by choosing a new variable $x = \sinh(y)$, the above integral can be rewritten in the following form $$ \int_0^\infty dp ~\cos\left(\sqrt{1+x^2}a\right) = \int_0^\infty dy \cosh(y) ~\cos\left(\cosh(y)a\right).~~~~(2) $$ This expression reminded me of one of the integral representations of the Bessel function $$ J_\nu (x) = \frac{2}{\pi} \int_0^\infty dy \cosh(\nu y) \sin\left(\cosh(y)x - \frac{1}{2}\nu \pi\right)~~~~(3) $$ where $x>0$ and $|\text{Re}(\nu)|<1$. The reference is taken from Chapter VI, Eq.(12) (page 180) of https://books.google.co.jp/books/about/A_Treatise_on_the_Theory_of_Bessel_Funct.html?id=Mlk3FrNoEVoC&redir_esc=y It is also in this website https://dlmf.nist.gov/10.9#i
This expression coincides with the above equation (2) if $\nu=1$. However, the range of the expression (3) does not include this point $|\text{Re}(\nu)|<1$.
Is there an expression for $J_1 (x)$ which takes similar form to the expression (3).