Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Per the title, do the integrals $\displaystyle\int_0^\infty \frac{\cos(\ln(x))}{x}\,dx$ and/or $\displaystyle\int_0^{\pi/2} \frac{\ln(\sin x)}{\sqrt{x}}\,dx$ converge?


I've no idea how to approach this. Dirchlet test doesn't tell me they converge/diverge, and the functions aren't nonnegative so I'm not sure if I can use the comparison test...

share|cite|improve this question
According to wolfram-alpha the second integral is approximately, $4.09808...$ – user17762 Dec 8 '11 at 3:59
up vote 2 down vote accepted

For the second integral, I would rewrite $\ln(\sin x)$ as $\ln(x)+\ln\left(\frac{\sin x}{x}\right)$. The second term in the numerator is bounded, and as for the first, note that $\frac{\ln x}{\sqrt x}<x^{-2/3}$ for sufficiently small $x$, which follows for example from l'Hôpital's rule.

share|cite|improve this answer
Using your hint I can show $ln(r)/\sqrt{r}$ exists using comparison. Am I right that for $ln(sinr/r)/\sqrt{r}$, the integral exists because the function is continuous/bounded in $(0,\pi/2]$, and so the sum of these two exists? – rol44 Dec 8 '11 at 4:20
@rol44: Right for the first part. For the second, not quite what you said; because $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$, it follows that $\ln\left(\frac{\sin x}{x}\right)$ is bounded on $\left(0,\frac{\pi}{2}\right)$, so the second integral can be compared to $\int\limits_0^{\frac{\pi}{2}}x^{-1/2}$. – Jonas Meyer Dec 8 '11 at 4:23
But since $ln(sinr/r)/\sqrt{r}$ is bounded in the same interval (since the limit at 0 is 0), wouldn't that guarantee the integral's existence without need for comparison? – rol44 Dec 8 '11 at 4:27
@rol44: Oh yeah, thanks, silly me. Come to think of it, comparing directly to $x^{-2/3}$ in the first place might even be easier. – Jonas Meyer Dec 8 '11 at 4:32
Ah, I see. That's probably the quickest way... Okay, I think this question is answered. I'm choosing Jonas's answer since it has more votes at this time, though both answers are equally useful. Thanks to both of you! – rol44 Dec 8 '11 at 4:34

For the first one, can you go on with the following indefinite integral $$ \int \frac{\cos(\ln(x))}{x}dx = \sin(\ln(x))? $$

Sivaram's comment may have given you an idea for the second one. (The goal is to show convergence.) I would like to do integration by part first for the indefinite integral, $$ \int\frac{\ln(\sin x) }{\sqrt x}dx = 2\int\ln(\sin x)d(\sqrt x) = 2\ln(\sin x)\sqrt{x}-2\int\sqrt {x}\frac{\cos x}{\sin x}dx $$ and then treat $\ln(\sin x)\sqrt{x}$ and $\int\sqrt {x}\frac{\cos x}{\sin x}dx$ separately.

share|cite|improve this answer
Oooh, awesome. Can't believe I missed that. Do you have a useful hint for the second integral? – rol44 Dec 8 '11 at 4:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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