In a 2010 exam question I found, no solution is provided. The question is:
Show that for all $K\in \mathbb{N} $ the integral converges:
$\int_{1}^{\infty}\frac{\cos^{2k+1}x}{x}dx$
I want to say that for every odd power of $\cos x$, since the positivity stays the same, the integral $\int \cos^{2k+1}xdx$ is bounded, and then use Dirichlet's Criterion.
I know how to actively divide the integral into the periodic regions, show that over every period the integral is 0 and bound between the maximum and minimum in the regions.
But is there a neater way to show this?
You can explicitly compute the integral $\int_{a}^{b}\cos^nxdx$. Define $J(n) = \int\cos^nxdx$ (indefinite integral). Then you get
$$J(n) = \sin x \frac{\cos^{n-1}x}{n} + \frac{n-1}{n}J(n-2) .$$
This gives a recursion ending with $J(1) = \sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = \sin x \cdot p_{n-1}(\cos x) + a_0x$$
where $p_{n-1}$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ \int_{1}^{b}\cos^{2k+1}xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $\int_{1}^{\infty}\frac{\cos^{2k+1}x}{x}dx$ converges. In fact, the sequence
$$a_m = \int_{m\pi - \pi/2}^{m\pi + \pi/2}\frac{\cos^{2k+1}x}{x}dx $$
converges to $0$ and the series
$$\Sigma_{m=1}^\infty a_m$$
is alternating and therefore converges.