<s>[INST] Find an asymptotic equivalent of the sequence $(\int_{-\infty}^{+\infty} \frac{1}{\cosh^n(x)} dx)_n$. 
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.
 [/INST] 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.
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