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My purpose is to show that this integral 

\begin{equation} I_t(x)=\int_{-\infty}^{\infty}e^{-\frac{\cosh^2(u)}{2x}}\,e^{-\frac{u^2}{2 t}}\,\cos\left(\frac{\pi\,u}{2t }\right)\,\cosh(u)\,du\,\,,\,\,x,t>0 \end{equation}

is positive ( $I_t(x)>0$ for all $x,t>0$). This integral can also be written as:

\begin{equation} I_t(x)=-e^{t/2}\,\int_{-\infty}^{\infty}e^{-\frac{\cosh^2(u+t)}{2x}}\,e^{-\frac{u^2}{2 t}}\,\sin\left(\frac{\pi\,u}{2t }\right)\,du \end{equation}

\begin{equation} I_t(x)=i\,e^{-\pi^2/8t}\,\int_{-\infty+i\,\pi/2}^{\infty+i\,\pi/2}e^{\frac{\sinh^2(u)}{2x}}\,e^{-\frac{u^2}{2 t}}\,\sinh\left(u\right)\,du \end{equation}

\begin{equation} I_t(x)=e^{-\pi^2/8t}e^{-1/2x}\,\int_{-\infty}^{\infty}e^{-\frac{u^2}{2x}}\,e^{-\frac{ArcCosh^2(i\,u)}{2 t}}\,du \end{equation}

It can also be used that $$f (x) = \frac {e ^ {\frac {\pi ^ 2} {8t}}} {2 \pi \sqrt {tx ^ 3}} \, I_t(x)$$

is a density function and thus $ \int _ {-\infty}^{\infty} f(x) dx = 1$

The way that I've chosen is splitting the integrand in periodic regions and try to see that positive areas are bigger than negative areas. Unfortunatelly , this not work (and believe me that I have tried a lot different variants !!!).

Attached the graphic of $I_t(x)$ for $t=1$ (for different values of t the graphic is similar): we can observe that function starts at 0 , grows up and finally decreases til 0 again.

enter image description here

Now I wonder how to prove the positivity by using derivatives of $I_t(x)$ with respect to $x$ (or $t$ !!). You think is possible?

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