Compute integral: $\lim_{n \to +\infty} \int_{1}^{+ \infty} \frac{\cos^{n}x}{x^2}dx$ I am pretty sure that the answer to $$\lim_{n \to +\infty} \int_{1}^{+ \infty} \frac{\cos^{n}x}{x^2}dx$$
is 0, but I am stuck in how to write a good proof, since I found it difficult to discuss the combination of two infinities.
I hope that you can help me solve the problem!
 A: Dominated convergence gives a one-liner proof. Since the integrand $x \mapsto \cos^n(x)/x^2$ is bounded by $1/x^2$ uniformly in $n$ and it converges to $0$ almost everywhere, we have
$$ \lim_{n\to\infty} \int_{1}^{\infty} \frac{\cos^n x}{x^2} \, \mathrm{d}x
= \int_{1}^{\infty} \lim_{n\to\infty} \frac{\cos^n x}{x^2} \, \mathrm{d}x
= \int_{1}^{\infty} 0 \, \mathrm{d}x
= 0. $$

If this fancy technique is not available yet, then still we can prove the claim. Let
$$ I_n = \int_{0}^{2\pi} |\cos^n x| \, \mathrm{d}x = 4 \int_{0}^{\pi/2} \cos^n x \, \mathrm{d}x.$$
It is not hard to show that $I_n \to 0$ as $n\to\infty$. This can be done in several ways, but here we discuss the following simple trick: substitute $\sin x = \frac{s}{\sqrt{n}}$. Then
$$ I_n = \frac{4}{\sqrt{n}} \int_{0}^{\sqrt{n}} \left( 1 - \frac{s^2}{n} \right)^{\frac{n-1}{2}}  \, \mathrm{d}s,$$
and using the inequality $1-x \leq e^{-x}$ which holds for all $x \in\mathbb{R}$,
$$ I_n \leq \frac{4}{\sqrt{n}} \int_{0}^{\sqrt{n}} e^{-\frac{n-1}{2n}s^2}  \, \mathrm{d}s. $$
Now it is clear that the integral $\int_{0}^{\sqrt{n}} e^{-\frac{n-1}{2n}s^2}  \, \mathrm{d}s$ is uniformly bounded in $n$, and so, this proves that $I_n \leq \text{const.} / \sqrt{n}$. So $I_n \to 0$.
Next, we bound the original integral using $I_n$ as follows:
\begin{align*}
\left| \int_{1}^{\infty} \frac{\cos^n x}{x^2} \, \mathrm{d}x \right|
&\leq \int_{1}^{\infty} \frac{|\cos^n x|}{x^2} \, \mathrm{d}x \\
&= \sum_{k=1}^{\infty} \int_{2\pi(k-1) + 1}^{2\pi k + 1} \frac{|\cos^n x|}{x^2} \, \mathrm{d}x \\
&\leq \sum_{k=1}^{\infty} \frac{1}{(2\pi(k-1) + 1)^2} \int_{2\pi(k-1) + 1}^{2\pi k + 1} |\cos^n x| \, \mathrm{d}x \\
&= C I_n.
\end{align*}
Here, $C = \sum_{k=1}^{\infty} \frac{1}{(2\pi(k-1) + 1)^2}$ is a finite constant. Now since $I_n \to 0$, it follows that the integral converges to $0$ as $n\to\infty$.
