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!
Best Answer
Help from @metamorphy. There seems to be a general strategy when solving a problem with some number to the power of a trigonometric function. Simply use y=trigonometric function (x) and substitute with the result usually being nothing more than a Bessel function or Struve function. For simplicity, I will not put down the series representation of the function as this one form works best. Using this source from the highlighted equation, one can use a Bickley function and simplify:
$$\mathrm{H(a)=\int_\Bbb R a^{cosh(x)}dx=2\int_0^\infty e^{-y\,cosh(t)}dt=2K_0(y)=\boxed{\mathrm{2K_0(-ln(a))}}\implies \int_\Bbb R cos(\pi cosh(x))dx=Re(H(-1))=-\pi Y_0(\pi)=-1.03959318…, H\left(\frac1e\right)=\int_\Bbb R e^{-cosh(x)}dx=2K_0(1)=\quad.8420488764…}$$
Seeing this plot, one can determine that the function named “H(a)” has the following behavior which still is satisfied by the original value of $\mathrm{H(a)=2K_0(-ln(a))}$. Still much is left to find on how this analytic continuation works.
$$\mathrm{0<a<1:H(a)=2K_0(ln(a^{-1})), a>1: H(a)=\int_0^\infty a^{-cosh(x)}dx=2K_0(ln(a))}$$
Here is some info on the Modified Bessel Function of the Second Kind. Just like the reference question, this creates an analytic continuation being “able” to take the divergent area for an “a” value and output a complex number. Here is an amazing complex plot of the result. All Bessel functions. Please correct me and give me feedback!