Is $\int_0^\infty dt\ \exp( - \cosh(t) z ) \sin(\nu t)$ related to Bessel functions?

Question

For $\nu>0$ and $z>0$ consider the integral $$ F(z, \nu) := \int_0^\infty dt\ e^{ - \cosh(t) z } \sin(\nu t) $$ It seems that this integral should be related to $I_{\nu}(z)$ according to some integral representations on DLMF, but after playing around with these integrals, I can't quite figure it out. Is there a way write an expression for $F(z, \nu)$?

Answer 1

Hint:

Consider $I_{i\nu}(z)=\dfrac{1}{\pi}\int_0^\pi e^{z\cos t}\cos i\nu t~dt-\dfrac{\sin i\nu\pi}{\pi}\int_0^\infty e^{-z\cosh t-i\nu t}~dt=\dfrac{1}{\pi}\int_0^\pi e^{z\cos t}\cosh\nu t~dt-\dfrac{i\sinh\nu\pi}{\pi}\int_0^\infty e^{-z\cosh t-i\nu t}~dt$

and $I_{-i\nu}(z)=\dfrac{1}{\pi}\int_0^\pi e^{z\cos t}\cos(-i\nu t)~dt-\dfrac{\sin(-i\nu\pi)}{\pi}\int_0^\infty e^{-z\cosh t+i\nu t}~dt=\dfrac{1}{\pi}\int_0^\pi e^{z\cos t}\cosh\nu t~dt+\dfrac{i\sinh\nu\pi}{\pi}\int_0^\infty e^{-z\cosh t+i\nu t}~dt$

$\therefore I_{i\nu}(z)+I_{-i\nu}(z)=\dfrac{2}{\pi}\int_0^\pi e^{z\cos t}\cosh\nu t~dt+\dfrac{i\sinh\nu\pi}{\pi}\int_0^\infty(e^{-z\cosh t+i\nu t}-e^{-z\cosh t-i\nu t})~dt=\dfrac{1}{\pi}\int_0^\pi e^{z\cos t+\nu t}~dt+\dfrac{1}{\pi}\int_0^\pi e^{z\cos t-\nu t}~dt-\dfrac{2\sinh\nu\pi}{\pi}\int_0^\infty e^{-z\cosh t}\sin\nu t~dt$