Show that the integrals converge or diverge:
1) $\int_\limits{1}^{\infty}\frac{\cos x}{x^{\alpha}}dx$ for $\alpha>\alpha_0>0$
2) $\int_\limits{1}^{\infty}\frac{\cos x}{x^{\alpha}}dx$ for $\alpha>0$
I thought of using Dirichlet's test on the first integral since $\lim_{x\to\infty}\frac{1}{x^{\alpha}}=0\forall \alpha$. However I think I would need to prove that $\int_\limits{1}^{\infty} \cos(x)$ converges uniformly.
Questions:
How should I do it? Am I on the right track?
What do you think of the second integral? What is the difference from the first?
Thanks in advance!
Convergence and, in fact, uniform convergence for case (1) follows from the Dirichlet test since $\displaystyle\left|\int_1^c \cos x \, dx \right| \leqslant2 $ for all $c > 1$ (uniformly bounded) and $x^{-\alpha} < x ^{-\alpha_0}$ which implies that $x^{-\alpha} \downarrow 0$ monotonically and uniformly for all $\alpha > \alpha_0$.
For case (2), we have convergence since the argument for case (1) applies to any $\alpha_0 > 0$. However, the convergence is not uniform.
Given sequences $\displaystyle c_n = -\frac{\pi}{4}+2\pi n$ and $\displaystyle d_n = \frac{\pi}{4}+ 2 \pi n$, we have $\cos x > 1/\sqrt{2}$ for $c_n \leqslant x \leqslant d_n$ and $$\left|\int_{c_n}^{d_n} \frac{\cos x}{x^\alpha} \right| \geqslant \frac{1}{d_n^\alpha}\int_{c_n}^{d_n} \cos x \, dx \geqslant \frac{1}{d_n^\alpha}\frac{\pi}{2 \sqrt{2}}.$$
Taking the sequence $\alpha_n = ( \log d_n)^{-1},$ we have $d_n^{\alpha_n} = \exp(\log d_n (\log d_n)^{-1})= e$ and, consequently,
$$\tag{*}\left|\int_{c_n}^{d_n} \frac{\cos x}{x^\alpha_n} \right| \geqslant \frac{\pi}{2 \sqrt{2}e}.$$
Since $c_n , d_n \to \infty$ and $\alpha_n \in (0,\infty)$ as $n \to \infty$, the Cauchy criterion for uniform convergence is violated. Note that uniform convergence would require that for any $\epsilon > 0$ there exists $K > 1$ such that for all $d> c> K$ and for any $\alpha \in (0,\infty)$ we have
$$\left|\int_{c}^{d} \frac{\cos x}{x^\alpha} \right| < \epsilon ,$$
which is contradicted by (*).