Laplace integral - Asymptotic expansion $$\int_{0}^{\infty} \frac{t^{x}}{{\cosh (t)}}dt$$
I'm trying to use Laplace's method to find the leading asymptotic behavior for x>>1, but I'm having some trouble. Could someone help me?
Thanks in advance,
 A: First note that 
$$\cosh t\geqslant\frac{\mathrm e^t}2,
$$ 
hence $I\leqslant J$ with
$$
J=2\int_0^\infty t^x\mathrm e^{-t}\,\mathrm dt=2\Gamma(x+1).
$$
Then use the bound 
$$\frac2{\mathrm e^t}-\frac1{\cosh t}\leqslant\frac2{\mathrm e^{3t}}
$$ 
to deduce that
$J-I\leqslant K$ with
$$
K=2\int_0^\infty t^x\mathrm e^{-3t}\,\mathrm dt=2\Gamma(x+1)3^{-(x+1)}.
$$
Thus, for every $x$,
$$
2\Gamma(x+1)(1-3^{-(x+1)})\leqslant I\leqslant2\Gamma(x+1),
$$
from which a simple equivalent should be clear. This also provides rigorous bounds for finite values of $x$, for example, if $x=4$,
$$
47.8024\approx2(4!)(1-243^{-1})\leqslant I\leqslant2(4!)=48,
$$
to be compared to the true value
$$
I=\frac{5\pi^5}{32}\approx47.8156.
$$
Note finally that, expanding $\frac1{\cosh t}$ as a power series in $\mathrm e^{-t}$, one gets the exact value
$$
I=2\Gamma(x+1)\sum_{n\geqslant0}\frac{(-1)^n}{(2n+1)^{x+1}}=2\Gamma(x+1)(1-3^{-(x+1)}+5^{-(x+1)}-7^{-(x+1)}+\ldots),
$$
which can be used as a refinement (and a confirmation) of the bounds above.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\int_{0}^{\infty}{t^{x} \over \cosh\pars{t}}\,\dd t
&=
2\int_{0}^{\infty}{t^{x}\expo{-t} \over 1 + \expo{-2t}}\,\dd t
=
2\int_{0}^{\infty}t^{x}\expo{-t}
\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\expo{-2\ell t}\,\dd t
\\[3mm]&=
2\sum_{\ell = 0}^{\infty}\pars{-1}^{\ell}\int_{0}^{\infty}t^{x}
\expo{-\pars{2\ell + 1}t}\,\dd t
=
2\sum_{\ell = 0}^{\infty}
{\pars{-1}^{\ell} \over \pars{2\ell + 1}^{x + 1}}
\overbrace{\int_{0}^{\infty}t^{x}\expo{-t}\,\dd t}^{\ds{\Gamma\pars{x + 1}}}
\end{align}

$$
\mbox{and}\quad
\sum_{\ell = 0}^{\infty}
{\pars{-1}^{\ell} \over \pars{2\ell + 1}^{x + 1}}
=
\beta\pars{x + 1}
$$

$$
\int_{0}^{\infty}{t^{x} \over \cosh\pars{t}}\,\dd t
=
2\beta\pars{x + 1}\Gamma\pars{x + 1}
$$
The asymptotic behavior, when $x \gg 1$, is found from the $\beta$ and $\Gamma$ features:
$$
\beta\pars{x + 1} \sim 1 - 3^{-x - 1}
\qquad\mbox{and}\qquad\Gamma\pars{x + 1} \sim \root{2\pi}x^{x + 1/2}\expo{-x}
$$
$$
\int_{0}^{\infty}{t^{x} \over \cosh\pars{t}}\,\dd t
\sim
2\root{2\pi}x^{x + 1/2}\expo{-x}\qquad\mbox{when}\qquad x \gg 1
$$
A: Write the integral as
$$
I=\int_{0}^{\infty}e^{A(t)}dt,
$$
where
$$
\begin{eqnarray}
A(t)&=&x\log t - \log\cosh t \\ &=&x\log t - t +\log 2- \log(1+e^{-2t}) \\ &\approx& x\log t-t+\log 2;
\end{eqnarray}
$$
we've dropped the last term, since it gives exponentially small contributions for large $t$.  The amplitude is largest at $t=x$, around which it can be expanded as
$$
A(t)=x\log x-x+\log2-\frac{1}{2x} (t-x)^2 + O(|t-x|^3).
$$
The integral can then be approximated by
$$
I\approx 2\left(\frac{x}{e}\right)^x\int_{-\infty}^{\infty}\exp\left(-\frac{1}{2x}(t-x)^2\right)dt=2\left(\frac{x}{e}\right)^x\sqrt{2\pi x}\approx 2(x!).
$$
This approximation works well even for small $x$; for instance, the exact result for $x=4$ is $5\pi^2/32\approx 47.8156$, which is quite close to $2(4!)=48$.
