I have an integral of the kind
$\int_{-\infty}^\infty e^{- d \cosh(x+i a)} dx $
where $d, a \in \mathbb{R}$. Now, I know that
$\int_{-\infty}^\infty e^{- d \cosh{x}} dx = 2 K_0(d)$
and I would like to express my integral in terms of bessel functions. However, if I deform the contour to obtain the second integral, the two vertical pieces will give me a contribution that doesn't vanish (it's possile to check this numerically as well, these two integral are not the same). Is there a way to express the first integral in terms of bessel functions? Or at least find an asymptotic form for the large $d$ behaviour?
No, the integral remains analytically and numerically the same as far as you keep $d$ real and positive and continuously deform $a$. However, it will become divergent as you will approach the lines $\Re{a}=\pm \frac{\pi}{2}$.
Actually the general answer (say, for $d\in\mathbb{R}_{>0}$) is:
$$\int_{-\infty}^{\infty}e^{-d\cosh(x+ia)}dx= \begin{cases} 2K_0(d), & \text{for } \Re{a}\in\left(-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n\right),\\ \text{divergent} & \text{for }\Re{a}\in\left(\;\frac{\pi}{2}+2\pi n\;,\frac{3\pi}{2}+2\pi n\right), \end{cases}$$ with $n\in\mathbb{Z}$ labeling different strips of convergence and divergence.