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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?

Attempt

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...

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According to wolfram-alpha the second integral is approximately, $4.09808...$ –  user17762 Dec 8 '11 at 3:59

2 Answers 2

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.

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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.

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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

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