I am having trouble with the following function, fix $n\in\mathbb{N}$, define $$f(x)=\int_0^\infty \int_0^\infty e^{10s-3e^{4s}\cosh(t)} s^{2n} \cos(xt) ds dt$$
Is this function positive for all $x\in\mathbb{R}$? I think it is, but am having trouble bounding the kernels. In particular, if it is negative, do we have bounds of where it can be negative? Does it only achieve being negative in one interval?
This is a partial result.
If we consider the inner more general integral $$I_n=\int_0^\infty e^{a s+b e^{c s} \cosh (t)}\,s^{n} \,\cos (x t )\,ds$$ Mathematica provides answers for specific values of $n$. Using $d=\frac{a+c}{c}$
$$I_1= +\frac{1! }{a^2}\cos (x t ) \,\, \, _2F_2\left(\frac{a}{c},\frac{a}{c};d,d;b \cosh (t)\right)$$ $$I_2= -\frac{2! }{a^3}\cos (x t ) \,\, \, _3F_3\left(\frac{a}{c},\frac{a}{c},\frac{a}{c};d,d,d;b \cosh (t)\right)$$ $$I_3=+\frac{3! }{a^4}\cos (x t )\,\, _4F_4\left(\frac{a}{c},\frac{a}{c},\frac{a}{c},\frac{a}{c};d,d,d,d;b \cosh (t)\right)$$ $$I_4=-\frac{4! }{a^5}\cos (x t )\,\,\, _5F_5\left(\frac{a}{c},\frac{a}{c},\frac{a}{c},\frac{a}{c},\frac{a }{c};d,d,d,d,d;b \cosh (t)\right)$$ and I suppose that we can conjecture $$I_n=(-1)^{n+1} \frac {n!}{a^{n+1}}\cos (x t )\,\,\,_{n+1}F_{n+1}\left(\frac{a}{c},\frac{a}{c},\cdots;d,d,\cdots;b \cosh (t)\right)$$ These integrals can also write in terms of the Meijer G-functions.
Now, I do not think that, even for fixed value of $(n,a,b,c)$, we could compute explicitly $$J_n=\int_0^\infty I_n\,dt$$
From a numerical point of view, the integration seems to be very difficult.
Trying for $J_2$ and $J_4$ with $a=10$, $b=-3$, $c=4$ for small velues of $x$, they seem to be negative (probably because of the $(-1)$ in front).