Different way to show $\int_{-\infty}^{\infty}e^{-\frac{\cosh^2(u)}{2x}}\,e^{-\frac{u^2}{2 t}}\,\cos\left(\frac{\pi\,u}{2t }\right)\,\cosh(u)\,du > 0$

I have been long time 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 , that is, $I_t(x)>0\,\,\forall x,t>0$ but I haven't had success.

Basically, I have used the following way: splitting the interval in differents regions where the integrand changes it sign and try to see that positive areas are bigger than negative areas. Unfortunately, this way not work (with a lot of different variants).

Now, at this moment, I wonder by another different way to prove it.

Anybody knows how to prove that this kind of function are positive in a different way? I only need the way, not the proof.

Attach the graphic of $I_t(x)$ for $t=1$ (is very similar for others values of $t$)

and add, if interesting, that

$$f (x) = \frac {e ^ {\frac {\pi ^ 2} {8t}}} {2 \pi \sqrt {tx ^ 3}} \, I_t(x)$$

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

Any help is welcome.

Solution 1:

Integral $I_t(x)$ 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} \tilde{I}_t(x)=e^{-1/2x}\int_{-\infty}^{\infty}e^{-\frac{u^2}{2x}}\,e^{-\frac{arcsinh^2(u)}{2 t}}\,\cos\left(\frac{\pi\,arcsinh(u)}{2t}\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}

\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}

In first three cases, I wonder if the integrand would be inferiorly bounded by a odd function, the proof is finished.

It's possible?