Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.

Question

Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$.

I found the result using a trick (using $e^x = \tan(\theta/2)$ as mentioned here) but I wanted to know if there isn't another less astute approach.

Answer 1

There were a variety of approaches presented on the page of the embedded link in the OP. Here is yet another way forward.

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Enforce the substitution $e^x\mapsto x$. Then, we see that

$$\begin{align} \int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx&=2^n\int_0^\infty \frac{1}{(x+x^{-1})^n}\frac1x \,dx\\\\ &=2^n\int_0^\infty \frac{x^{n-1}}{(x^2+1)^n}\,dx\tag1 \end{align}$$

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Next, enforcing the substitution $x\mapsto x^{1/2}$ in $(1)$ reveals

$$\begin{align} 2^n\int_0^\infty \frac{x^{n-1}}{(x^2+1)^n}\,dx&=2^{n-1}\int_0^\infty \frac{x^{n/2-1}}{(x+1)^n}\,dx\\\\ &=2^{n-1}B\left(\frac{n}{2},\frac{n}{2}\right)\\\\ &=B\left(\frac{n}{2},\frac{1}{2}\right)\tag2 \end{align}$$

where $B(x,y)$ is the Beta Function. Note that the result in $(2)$ agrees with the result reported on the referenced page.

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The equivalence of $2^{n-1}B\left(\frac n2,\frac n2\right)=B\left(\frac n2,\frac12\right)$ can be established using the relationship between the Beta Function and the Gamma Function, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ along with Legendre's Duplication Formula of the Gamma Function, $\Gamma (z)\Gamma (z+1/2)=2^{1-2z}\Gamma(1/2)\Gamma(2z)$ with $z=n/2$.

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For large $n$, we can use the relationship between the Beta Function and the Gamma Function, $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$, along with Stirling's Approximation to arrive at the asymptotic approximation

$$B\left(\frac{n}{2},\frac{1}{2}\right)\sim\sqrt{\frac{2\pi}{n}}$$

Therefore, we see that for large $n$ the integral of interest has the asymptotic approximation

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\sim \sqrt{\frac{2\pi}{n}}}$$

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APPENDIX: SIMPLE ESTIMATES TO OBTAIN THE ASYMPTOTIC EXPANSION

It is straightforward to show that the hyperbolic cosine satisfies the inequalities

$$1+\frac12x^2\le \cosh(x)\le e^{x^2/2}\tag{A1}$$

Hence, we have from $(A1)$

$$\int_{-\infty}^\infty e^{-nx^2/2}\,dx \le\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\le \int_{-\infty}^\infty \frac{1}{\left(1+\frac12 x^2\right)^n}\,dx \tag{A2}$$

The value of the integral on the left-hand side of $(A2)$ is $\sqrt{\frac{2\pi}n}$ while the value of the integral on the right-hand side of $(A2)$ can be found by making the substitution $x\mapsto \sqrt 2\tan(x)$. To wit, we see that

$$\int_{-\infty}^\infty \frac1{\left(1+\frac12 x^2\right)^n}\,dx=2\sqrt{2}\int_0^{\pi/2}\cos^{2n-2}(x)\,dx\tag{A3}$$

The integral on the right-hand side of $(A3)$ is Wallis' Integral with exponent $2n-2$.

It is straightforward to show (See Here) that $\lim_{n\to \infty}\sqrt{n}\int_0^{\pi/2}\cos^{2n-2}(x)\,dx=\frac{\sqrt{\pi}}{2}$ whence application of the squeeze theorem reveals

$$\lim_{n\to\infty}\sqrt{n}\int_{-\infty}^\infty \frac1{\cosh^n(x)}\,dx=\sqrt{2\pi}$$

from which we find the asymptotic approximation

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1}{\cosh^n(x)}\,dx\sim \sqrt{\frac{2\pi}{n}}}$$

as expected!

Answer 2

There is a different approach: You can derive a recursive relation for $a_n:=\int_{\mathbb{R}} \cosh^{-n}(x) dx$ using the identity $\cosh^2-\sinh^2=1$ and integration by parts. In fact, for $n\geq 3$, we have

$a_n= \frac{n-2}{n-1}a_{n-2}$.

Indeed,

$a_n=\int_{\mathbb{R}} \frac{1}{\cosh^{n}(x)} dx= \int_{\mathbb{R}} \frac{\cosh^2(x)-\sinh^2(x)}{\cosh^{n}(x)} dx= \int_{\mathbb{R}} \frac{1}{\cosh^{n-2}(x)}dx -\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx=a_{n-2}-\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx.$

Now, integration by parts shows that the remaining integral is equal to $\frac{1}{n-1}a_{n-2}$. Indeed, recalling the identites $\cosh'=\sinh$ and $\sinh'=\cosh$, we have

$\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx = \int_{\mathbb{R}}\frac{\sinh(x)}{\cosh^{n}(x)} \cdot \sinh(x) dx= \left[-\frac{1}{n-1}\cosh^{-(n-1)}(x)\cdot \sinh(x)\right]_{-\infty}^{\infty}$

$+ \int_{\mathbb{R}} \frac{1}{n-1}\cosh^{-(n-1)}(x) \cdot \cosh(x) dx = 0 + \frac{1}{n-1}a_{n-2}=\frac{1}{n-1}a_{n-2}$.

So,

$a_n=a_{n-2}-\int_{\mathbb{R}}\frac{\sinh^2(x)}{\cosh^{n}(x)}dx = a_{n-2}-\frac{1}{n-1}a_{n-2}=\frac{n-2}{n-1}a_{n-2}$.

By induction, it follows that

$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{2}{3}a_2$ for $n$ even.

and

$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{1}{2}a_1$ for $n$ odd.

Therefore, it remains to compute $a_1$ and $a_2$. Now, as $\cosh(x)=\frac{1}{2}(e^x+e^{-x})$, we have

$a_1=2\int_{\mathbb{R}} \frac{1}{e^x+e^{-x}} dx = 2\int_{\mathbb{R}} \frac{e^x}{(e^x)^2+1} dx =2 \int_{\mathbb{R}} \frac{1}{y^2+1} dy = 2 [\arctan(y)]_{-\infty}^{\infty}=2\pi.$

Similarly, one can show $a_2=2$. Hence,

$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{2}{3}\cdot 2$ for $n$ even.

and

$a_{n}=\frac{n-2}{n-1}\frac{n-4}{n-3}\dots \frac{1}{2}2 \pi$ for $n$ odd.

Answer 3

Another way to show the integral is asymptotic to $\sqrt{2\pi/n}$ is to use the small-$x$ approximation $\ln\cosh x\approx\tfrac12x^2+\frac{1}{24}x^4$ to approximate the integrand near the $y$-axis as $\exp\tfrac{-nx^2}{2}$, whose integral is $\sqrt{2\pi/n}$. There is little contribution from $x$ of large modulus, for which $\cosh x\approx\tfrac12e^{|x|}$ so the integrand is asymptotic to $2^ne^{-n|x|}$. Indeed, we expect the small-$x$ approximation to loose its robustness at some positive $x\in O(1)$, but $\int_1^n2^ne^{-nx}dx=(2/e)^n/n$ is small compared with our integral approximation.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} {\dd x \over \cosh^{n}\pars{x}}} = 2\int_{0}^{\infty} \expo{-n\ln\pars{\cosh\pars{x}}}\,\,\,\,\dd x \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim} \,\,\,\,\,&\ 2\int_{0}^{\infty} \exp\pars{-{nx^{2} \over 2}}\,\,\,\,\dd x = 2\root{2 \over n}\int_{0}^{\infty} \expo{-x^{2}}\,\,\,\,\dd x \\[5mm] = & 2\root{2 \over n}{\root{\pi} \over 2} = \bbx{\root{2\pi} \over n^{1/2}} \\ & \end{align}