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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!

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  • $\begingroup$ To apply Abel's test, you need only show that $\frac1{x^\alpha }$ monotonically decreases to $0$ and that there exists a number $C$, such that for every $L$, $\int_1^L \cos(x)\,dx\le C$. $\endgroup$ – Mark Viola May 20 '18 at 18:23
  • $\begingroup$ Your limit is not correct... $\frac{1}{x^\alpha}$ does not even depend on $n$. Perhaps you mean the limit as $\alpha \to \infty$? Of course, then you need to note that $x > 1$, as then the limit goes to $1$ and not $0$ $\endgroup$ – Brevan Ellefsen May 20 '18 at 19:13
  • $\begingroup$ @BrevanEllefsen I am referring to $x\to\infty$. Please check my update! Thanks for finding the mistake! $\endgroup$ – Pedro Gomes May 21 '18 at 13:45
  • $\begingroup$ Shouldn't the question be does the improper integral converge uniformly for (1) $\alpha \in [\alpha_0,\infty)$ and (2) $\alpha \in (0,\infty)$? Otherwise as you ask -- what is the difference in the two cases? $\endgroup$ – RRL May 21 '18 at 19:59
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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 (*).

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