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.

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    $\begingroup$ I did NIntegrate[Cosh[x]^(-x), {x, 0, Infinity}] in the Alpha cloud and got $1.22494$, so that gives you an idea how good the approximation is. $\endgroup$ – Ross Millikan Dec 9 '16 at 16:17
  • $\begingroup$ Many thanks for your useful remarks and help, I fixed the typo, I hope that now there are no more mistakes. Thanks @RossMillikan $\endgroup$ – user243301 Dec 9 '16 at 16:19
  • $\begingroup$ Thanks for your edit, and attention @AntonioVargas $\endgroup$ – user243301 Dec 10 '16 at 19:55

Consider $$f(x)=\cosh ^{-x}(x)-2^x e^{-x^2}$$ and its values for some $x$ $$f(0)=0$$ $$f(1)=\text{sech}(1)-\frac{2}{e}\approx -0.0877046 $$ $$f(2)=\text{sech}^2(2)-\frac{4}{e^4}\approx -0.00261173$$ $$f(3)=\text{sech}^3(3)-\frac{8}{e^9}\approx -7.30541\times 10^{-6}$$ Let us use trapezoidal integration to get something better.


Let us consider $$I_a=\int_a^\infty \cosh ^{-x}(x)\,dx \qquad J_a=\int_a^\infty 2^x e^{-x^2}\,dx$$ The differences are larger when $a$ is close to $0$ as shown below $$\left( \begin{array}{ccc} a & I_a & J_a \\ 0.00 & 1.22494 & 1.37504 \\ 0.25 & 0.97543 & 1.10789 \\ 0.50 & 0.73242 & 0.82767 \\ 0.75 & 0.51027 & 0.56794 \\ 1.00 & 0.32539 & 0.35521 \\ 1.25 & 0.18796 & 0.20124 \\ 1.50 & 0.09766 & 0.10278 \\ 1.75 & 0.04542 & 0.04714 \\ 2.00 & 0.01885 & 0.01936 \end{array} \right)$$

Now, let us focuse on the function $$h(a)=2^a e^{-a^2}-\cosh ^{-a}(a)$$ that we can evaluate at different values of $a$ for, say, $0 \leq a \leq 2$ (step size=$0.1$) and for which we can build an interpolating function and integrate from $0$ to $2$. The result is $0.149594$ and $1.37504-0.14959=1.22545$$ which is very close to the value reported by Ross Millikan.

  • $\begingroup$ Many thanks, I had doubts, but now I understand your calculations, then the way is consider the differnce, that is the first line and use one rule of integration. Many thanks. $\endgroup$ – user243301 Dec 9 '16 at 22:18

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