I am trying to evaluate the integral $$I = \int_{-\infty}^{\infty} \underbrace{\coth x}_{f(x)} \exp(-a \cosh x + b \sinh x) \, dx \, ,$$ where $a \in \mathbb{R}^+$, $b \in \mathbb{R}$, and $|b/a| < 1$.
Related integrals, where one replaces $f(x) = \coth x$ by $\cosh^n x \sinh^m x$ for $n, m \in \mathbb{N}$, are relatively straightforward to solve by repeatedly differentiating under the integral, i.e. $$\int_{-\infty}^{\infty} \cosh^n x \sinh^m x \exp(-a \cosh x + b \sinh x) \, dx = (-\partial_a)^n \partial_b^m \int_{-\infty}^{\infty} \exp(-a \cosh x + b \sinh x) \, dx \, .$$ This integral can be evaluated to get a modified Bessel function of the second kind $$\int_{-\infty}^{\infty} \exp(-a \cosh x + b \sinh x) \, dx = 2 K_0(\sqrt{a^2 - b^2}) \, ,$$ whereupon one may straightforwardly apply the derivatives to generate the even $(\cosh^n x)$ and odd $(\sinh^m x)$ moments of the distribution.
However, this same approach does not work for the negative integer moments, since one must integrate for negative powers, with the integral in question being rather non-trivial to evaluate. For example, in the case that $f(x) = 1/\sinh x$, one finds $$I'(b) = \int_{-\infty}^{\infty} \exp(-a \cosh x + b \sinh x) \, dx \, .$$ Then, using the fact that the integrand of $I(0)$ is odd, such that $I(0) = 0$, one must find $$I = \int_0^b \int_{-\infty}^{\infty} \exp(-a \cosh x + b' \sinh x) \, dx \, db' = \int_0^b 2 K_0(\sqrt{a^2 - b'^2}) \, db' \, ,$$ and I do not know how to proceed. Similar difficulties are encountered for $f(x) = 1/\cosh x$. If one wants to evaluate the original integral by this method, with $f(x) = \coth x$ (or also for $f(x) = \tanh x$), then one must evaluate the above cases.
I would greatly appreciate any suggestions on how to evaluate the above integrals (for $f(x) = 1/\sinh x$ and $f(x) = 1/\cosh x$), or any suggestions of alternative approaches that would avoid these difficulties.