After the success and great answer by @metamorphy of my
question, I experimented a bit more and found this nice graph. This looks almost like a gamma function integral. The summed integral seems to diverge for. This also used a bit of inverse function magic. Many series for the inverse hyperbolic cosine require the argument$\ \in (-1,1)$. The values of the integral exist even for $a=-1$, $i$, so this value for $a$ would be amazing for finding these results.
$$H(a)\mathop=^\text{def}\mathrm{\int_{-\infty}^\infty a^{cosh(x)}dx=2\int_0^\infty\sqrt{a^{e^{-x}}a^{e^x}}dx=-2ln(a)\int_1^\infty a^t cosh^{-1}(t)dx \\ =2\sum_{n=0}^\infty\frac{ln^n(a)}{n!}\int_0^\infty cosh^n(x)dx=2\int_1^\infty\frac{a^t}{\sqrt{t^2-1}}dt}$$
A non-integral representation is wanted, but a closed form is optional. I will keep working on this problem. Please correct me and give me feedback!