I have been several days trying to prove that the following integral
$$I_t(x)=\int_{-\infty}^{\infty}e^{-\frac{\cosh^2(u)}{2x}}\,e^{-\frac{u^2}{2 t}}\,\cos\left(\frac{\pi\,u}{2t }\right)\,\cosh(u)\,du,\hspace{0.5cm}x,t>0$$
is positive ( $I_t(x)>0$ for all $x,t>0$) but I can't do it. Always, the way I choose is splitting the integrand in periodic regions and try to see that positive areas are bigger than negative areas. But, at least for me, this way not work. And believe me that I have tried a lot different variants !!!
So, maybe I should try to consider the function as such (not as a sum of areas) and look for something with the derivatives with respect to x or even with respect to t.
I also tried to prove that $I_t ^ 2 (x)>0$ trying to apply something in the resulting double integral but I think is not a good idea.
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 $\int _ {- \infty} ^ {\infty} f (x) dx = 1$.
Please lend a hand. Any help is welcome.
Since you only want to prove the integral is greater than $0$, I'll sketch a mode of proof. Consider the functions that comprise the integrand. $e^x$ is never less than $0$ in the real plane. $\cosh(x)$ is never less than $0$ in the real plane. $\cos(x)$ is your only real problem. Here's the thing to note, $cosh(x)$ increases slower than $e^{-cosh(x)^2}$ decreases. Since the function starts at 1 and ends up at $0$ on both sides of the interval, and since the integral of $/cos(x)$ is periodic and spends equal amounts of time + and -, the integral must be positive as the function decreases from 1 to 0 and spends equal amounts of time + and - after a certain point in the interval, which results in a integral of 0. Since only the beginning contributes more to the integral, and because its positive, the integral must be positive.