How to evaluate $\int_0^ \infty e^{-x\sinh(t)-\frac{1}{2}t}~dt$? $$
\int_0^ \infty e^{-x\sinh(t)-\frac{1}{2}t}~dt
$$
I tried doing it by parts and looking for differentials but I just keep getting back to the original expression. I can't think of a clever substitution either. Mathematica is giving me a complex answer with special functions:
$$
\frac{e^{-ix}\sqrt{\frac{\pi}{2}}(-i+ie^{2ix}\text{Erfc}[(-1)^{\frac{1}{4}}\sqrt{x}]+\text{Erfi}[(-1)^\frac{1}{4}\sqrt{x}])}{\sqrt{x}} 
$$
For real $x>0$, it does evaluate to real answers though.
 A: You can have a closed form in terms of the AngerJ function

$$ \frac{2}{x}+\sqrt{2\pi}\,\frac{\cos(x)}{x^{3/2}}+\sqrt{2\pi}\frac{\sin(x)}{x^{1/2}}-\pi {\bf{J}}_{\frac{1}{2}} ( x )-\frac{\pi}{x}{ \bf{J}}_{-\frac{1}{2}} ( x ) .$$

A: $\int_0^\infty e^{-x\sinh t-\frac{t}{2}}~dt$
$=2\int_\infty^0e^\frac{x(e^{-t}-e^t)}{2}~d(e^{-\frac{t}{2}})$
$=2\int_0^1e^{\frac{xt^2}{2}-\frac{x}{2t^2}}~dt$
$=\left[\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{i\sqrt xt}{\sqrt2}+\dfrac{\sqrt x}{\sqrt2t}\right)+e^{-ix}~\text{erfc}\left(\dfrac{i\sqrt xt}{\sqrt2}-\dfrac{\sqrt x}{\sqrt2t}\right)\right)\right]_0^1$ (according to http://dlmf.nist.gov/7.7#E7)
$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}\left(\text{erfc}\left(\dfrac{(i+1)\sqrt x}{\sqrt2}\right)-\text{erfc}(+\infty)\right)+e^{-ix}\left(\text{erfc}\left(\dfrac{(i-1)\sqrt x}{\sqrt2}\right)-\text{erfc}(-\infty)\right)\right)$
$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{(i+1)\sqrt x}{\sqrt2}\right)+e^{-ix}\left(\text{erfc}\left(\dfrac{(i-1)\sqrt x}{\sqrt2}\right)-2\right)\right)$
$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{(1+i)\sqrt x}{\sqrt2}\right)-e^{-ix}~\text{erfc}\left(\dfrac{(1-i)\sqrt x}{\sqrt2}\right)\right)$
