Prove that $\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$ converges

Question

Prove the convergence of

$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$

First I thought the integral does not converge because

$$\int\limits_1^{\infty} -\frac{1}{x} \,\mathrm{d}x \le \int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$$

But in this case

$$\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x \le \int\limits_1^{\infty} \frac{1}{x^2} \, \mathrm{d}x$$

it converges concerning the majorant criterion. What's the right way?

If you want to use the (direct) comparison test(s), be sure to check if you can apply the test (more specifically, look for conditions on the sign of the functions). — StackTD, Jul 15 '16 at 12:28
@jacmeird The problem is that $\frac{\cos x}{x}\leq \frac{1}{x^2}$ is not true! Take any $x>1$ such that $\cos x = 1$ and you will see. — 5xum, Jul 15 '16 at 12:33
@5xum Ouch, of course you're right. If I take $\frac{cos(x)}{x} \le \frac{1}{x}$, there is the same problem. — jacmeird, Jul 15 '16 at 12:35
You can apply most, if not all, of the same techniques found here — , Jul 15 '16 at 12:38

Olivier Oloa · Accepted Answer · 2019-01-11T04:38:32.403

5

You might want to use integration by parts, obtaining for $M\ge1$, $$ \int_{1}^M \frac{\cos x}{x}\: dx=\left[\frac{\sin x}{ x}\right]_1^M+ \int_1^M \frac{\sin x}{x^2}\: dx $$ letting $M \to \infty$ gives $$ \int_{1}^\infty \frac{\cos x}{x} \:dx=\lim_{M \to \infty}\int_1^M \frac{\cos x}{x} \:dx= -\sin 1+\int_1^\infty \frac{\sin x}{ x^2}\: dx $$then one may conclude by the absolute convergence of the latter integral: $$ \left|\int_1^\infty \frac{\sin x}{ x^2}\: dx\right|<\int_1^\infty \frac{|\sin x|}{ x^2}\: dx<\int_1^\infty \frac{1}{x^2}\: dx<\infty. $$

edited Jan 11 '19 at 04:38

answered Jul 15 '16 at 12:36

Olivier Oloa

120,989

Is it $\left|\lim_{M \to \infty}\int_{1}^M \frac{\cos x}{x} dx\right|$ ? – Mc Cheng Jul 15 '16 at 12:43
@McCheng Edited. Thank you very much! – Olivier Oloa Jul 15 '16 at 12:44
1

You should leave off those absolute values in your last line. – zhw. Jul 15 '16 at 15:56
@zhw. You are right, done. – Olivier Oloa Jan 09 '19 at 08:24
I think you lost a minus sign in front of $\frac{\sin 1}{1}$, right? – Ixion Jan 09 '19 at 10:48
@Ixion You are right, please feel free to edit my answer. Thank you. – Olivier Oloa Jan 09 '19 at 18:32

score 0 · Answer 2 · edited Jan 09 '19 at 09:01

Hint: Awful and tricky but as the problem is near $+\infty$, write $$\int \limits_{\frac{\pi}{2}}^{N\pi+\frac{\pi}{2}}\frac{\cos(x)}{x}\mathrm{d}x= \sum \limits_{k=1}^{N}\left(\int\limits_{k\pi-\frac{\pi}{2}}^{k\pi+\frac{\pi}{2}}\frac{\cos(x)}{x}\mathrm{d}x\right)$$ and use the criterion for alternating series with the sequence $$a_k=\int\limits_{k\pi-\frac{\pi}{2}}^{k\pi+\frac{\pi}{2}}\frac{\cos(x)}{x}\mathrm{d}x$$

score 0 · Answer 3 · answered Jan 09 '19 at 10:59

$$ \begin{align} \left|\,\int_1^\infty\frac{\cos(x)}{x}\,\mathrm{d}x\,\right| &\le\left|\,\int_1^\pi\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|+\sum_{k=1}^\infty\left|\,\int_{(2k-1)\pi}^{(2k+1)\pi}\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|\\ &=\left|\,\int_1^\pi\frac{\cos(x)}{x}\,\mathrm{d}x\,\right|+\sum_{k=1}^\infty\left|\,\int_{(2k-1)\pi}^{(2k+1)\pi}\cos(x)\left(\frac1{x}-\frac1{2k\pi}\right)\,\mathrm{d}x\,\right|\\ &\le\log(\pi)+\sum_{k=1}^\infty2\pi\left(\frac1{(2k-1)\pi}-\frac1{2k\pi}\right)\\[6pt] &=\log(4\pi) \end{align} $$

Prove that $\int\limits_1^{\infty} \frac{\cos(x)}{x} \, \mathrm{d}x$ converges

3 Answers3