Integral involving K modified Bessel function with respect to order I'm looking for to solve a definite integral involving  the modified Bessel function $K_{1/2+i\,u}(x)$.
The problem is that this integral is with respect to the order:
$$\int_{-\infty}^{\infty}e^{-2ty^2}\cosh(\pi\,y)\,K_{1/2+i\,y}(x)\,dy$$
with $t,x\geq0$. This integral is a real value integral. Using the well known relation between $I_u(x)$ and $K_u(x)$, we also have
$$\int_{-\infty}^{\infty}e^{-2ty^2}\left(I_{\frac{-1}{2}-i\,y}(x)-I_{\frac{1}{2}+i\,y}(x)\right)\,dy$$
I had a look in the classical books of integrals but I haven't found what I need. Probably, this integral can't be expressed in terms of elementary functions but I would like to know some about it: monotony, sign, ...
 A: Here is an alternative form of the integral that is good because (1) the integrand is composed of simple functions so numerical analysis is easy and (2) asymptotic analysis is made easy.  The final result is
$$ J := \int_{-\infty}^\infty e^{-2\,t\,y^2}\cosh{\pi y} \, K_{1/2+iy}(x) \, dy
$$ 
$$ =\frac{1}{2}\sqrt{\frac{\pi}{2t}}\exp{\big(\frac{\pi^2}{8t}\big)} \int_{-\infty}^\infty \exp{\big(-x\,\cosh(u) -\frac{u^2}{8t}-\frac{u}{2}\big)}
\cos{(\frac{\pi\, u}{4t} )}\, du. \, \, \, \, (1)
$$
The proof consists of inserting the for the McDonald function
$$K_v(z)=\int_0^\infty e^{-z\,\cosh{u}}\,\cosh(v\,u)\,du,$$
interchanging the integration, and performing the inner one as a Gaussian integral.  By extending the integral from $(0,\infty)$ to  $(-\infty,\infty)$ one gets the nice form presented.  Now let's do some simple asymptotic analysis.  For $x \to \infty$ the integrand in (1) will be well approximated by keeping the first two terms in $\cosh(u).$  The resulting integrand is again of Gaussian form so we find that, using $\kappa= 1/(8t),$
$$ J \sim \frac{\pi}{2} \big(t(2\kappa+x)\big)^{-1/2} \exp{\big(\kappa\,\pi^2-x+\frac{1-(4\kappa\pi)^2}{16\kappa+8x} \big)}\,\cos{\big(\frac{\kappa\pi}{2\kappa+x}\big)}\, \, , \, x\to\infty$$
To find out when this is a good approximation, use the approximation 
$$\exp{(-x\,\cosh(u))} \sim \exp{(-x\,(1+u^2/2))}\big(1-\frac{x\,u^4}{24} \big). $$  The second term, found explicitly as the fourth moment of a Gaussian, needs to be small, say, less that 1/10 the first term.  This gives you a way to find which $t$ are appropriate.  Note that the asymptotic expression already answers one of your questions: $J$ is not monotonic.  For $x$ very small, just use the Taylor series expansion and again you end up with integrals with Gaussian integrands.  Numerical analysis with $t\approx 1/2$ shows about 3 significant figures agreement for $x$ as small as 10.$  
