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I would like to determine if the following converges, and if possible compute it:

$$\int^\infty_{2/\pi}\ln\left(\sin{1\over x}\right)\,\mathrm dx$$

A bit of thought and a glance at a computer generated graph reveals that this is a horrific function. I performed a substitution to obtain:

$$\int_0^{\pi/2}\frac{\ln(\sin x)}{x^2}\,\mathrm dx$$

Which seems a bit more civilized. The integrand is continuous at $\frac{\pi}{2}$, so no problems there. I was hoping to show that the integrand is bounded by some integrable function like $\frac{1}{x^{1/2}}$, by e.g. multiplying by $x^{1/2}$ and taking a limit, but I can't see how to take a limit of $x^{1/2}\ln(\sin x)$.

Milind Hegde
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Jack M
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    The integral under consideration diverges because $\sin \frac 1 x \sim \frac 1 x, , x \to \infty$. Therefore, $\ln (\sin (\frac 1 x )) \sim - \ln (x),, x \to \infty$. – user64494 Oct 04 '13 at 17:52
  • @user64494 Could you make that an answer, with a few words on how to prove that $\sin\frac{1}{x}\sim\frac{1}{x}$? – Jack M Oct 04 '13 at 18:16
  • @ Jack M: The relation $\sin \frac 1 x \sim \frac 1 x, , x \to \infty $ follows from the limit $\lim_{t \to 0} \frac {\sin(t)} t =1 $ (see Wiki ) and the chain rule (ibid). – user64494 Oct 04 '13 at 19:31

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For $t\in(0,\pi/2]$, we have $\sin t<t$. This can be seen using the Taylor expansion of the sine and Leibnitz' criterion for convergence of alternating series; or simply note that $\sin t$ and $t$ have the same derivative at $t=0$ and that of $\sin t$ inreases less than $t$ does.

So, for $x\geq 2/\pi$, $$ \sin\frac1x<\frac1x. $$ As the logarithm is monotone, $\ln(\sin\frac1x)<\ln(\frac1x)=-\ln x$ for all $x\geq2/\pi$. Then $$ \int_{2/\pi}^\infty\,\ln(\sin\frac1x))\,dx\leq-\int_{2/\pi}^\infty\,\ln x\,dx=-\infty. $$ So the integral diverges.

Martin Argerami
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