Laplace integral - Asymptotic expansion

Question

$$\int_{0}^{\infty} \frac{t^{x}}{{\cosh (t)}}dt$$

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,

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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: $$ \beta\pars{x + 1} \sim 1 - 3^{-x - 1} \qquad\mbox{and}\qquad\Gamma\pars{x + 1} \sim \root{2\pi}x^{x + 1/2}\expo{-x} $$ $$ \int_{0}^{\infty}{t^{x} \over \cosh\pars{t}}\,\dd t \sim 2\root{2\pi}x^{x + 1/2}\expo{-x}\qquad\mbox{when}\qquad x \gg 1 $$

Answer 3

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 $2(4!)=48$.