Since it's easy to show that $$\frac{(2e^{x})^x}{(e^{2x}+1)^x}\leq \frac{2^x}{e^{x^2}}$$ for $x\geq 0$, I say using $e^{2x}+1\geq e^{2x}$, then one can calculate an approximation for $$\int_0^{\infty}\left(\frac{1}{\cosh x}\right)^xdx,$$ since (I've used Wolfram Alpha) one can use the definition of the error function to get a closed form of $$\int_0^\infty\frac{2^x}{e^{x^2}}dx.$$
That is $\approx 1.37504$ (see the closed-form, and thus the exact value in terms of the error function, with this code integrate 2^x/e^(x^2) dx, from x=0 to infnite).
I try do more calclations with $\int_0^{100}$, $\int_0^{1000}$ and integrand the genuine, to know if such approximation was good.
Question. Is it possible improve previous calculation to get a better appoximation of our integral $$\int_0^{\infty}\left(\frac{1}{\cosh x}\right)^xdx?$$ You can deduce your result with analysis, or well computing using numerical analysis, but in this case if it is possible explain us how/why works your method. Many thanks.
Were fixed some mistakes and typos, thanks for the patience, and help.