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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...$2011-12-08

2 Answers 2

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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} for sufficiently small $x$, which follows for example from l'Hôpital's rule.

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    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!2011-12-08
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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. Ca$n$'t believe I missed that. Do you have a useful hint for the second integral?2011-12-08