Is there some way to express:


$$I(t) = \int\_{-\infty}^{t} e^{-2\mathrm{cosh}(x)}~\mathrm{d}x$$


From Bessel functions?


By substituting $y = \mathrm{cosh}(x)$ we get 


$$I(t) = \int\_{1}^{\mathrm{cosh}(t)} \frac{e^{-2 y}}{\sqrt{y^2-1}}~\mathrm{d}y$$


In this form, Mathematica will give $I(\infty) = \mathrm{BesseK}(0,2)$ but not indication of what $I(t)$ might be. 


I can write the integral as:


$$I(t) = \int\_{1}^{\mathrm{sinh}(t)} \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}~\mathrm{d}z$$


and try to expand $f(x) = \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}$ as


$$f(z) = \sum\_{i=0}^{\infty} (-1)^j\frac{2~\mathrm{BesselK}\left(\frac{1}{2}+j,2\right)}{j!\sqrt{\pi}} z^{2j}$$


Unfortunately the series diverges for $z \ge 1$, as does the series for 
$f(z) z e^{2 z}$



Integral of exp(-2cosh(x))
By changing $s=e^{t-x}$ in the integral, one obtains $$I(t)=\int\_1^\infty e^{-e^{-t}s-e^t/s}\frac{ds}{s}$$ which corresponds to the definition of the incomplete Bessel function in [Harris](https://www.researchgate.net/publication/242980699_Incomplete_Bessel_generalized_incomplete_gamma_or_leaky_aquifer_functions): 
$$ I(t)=K\_0(e^{-t},e^t)$$ With eq. (38) in the article,
$$I(t)=K\_0(2)+tI\_0(2)+2\sum\_{j=1}^\infty \frac{(-1)^j}{j}I\_j(2)\sinh(jt)$$
where $I\_n$ and $K\_n$ are the modified Bessel functions. The author mentions that the expansion is absolutely convergent. 


