1
$\begingroup$

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!

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 (*).

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.