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Could you help me how to find the limit of $$\left(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1\right)^{\frac{1}{n}}?$$

I know that $$\ln \left((\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}} \right)=\frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right)$$

and $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( \sin(\frac{k}{n})\right) = \int_0^1 \ln(\sin(x)) \, dx \text{ (Riemann integral)}$$

but I am not sure what to do next, I mean, how do I get back to $$\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1)^{\frac{1}{n}}?$$

Could you help me with that?

Michael Hardy
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Hagrid
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    Just exponentiate the result of the integration. – Ron Gordon Jun 04 '13 at 17:49
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    And be carefull. $\int_0^1 \ln(\sin(x)) dx$ is not a Riemann Intregral. It is an improper integral. See discussions here: http://math.stackexchange.com/questions/406819/find-the-limit-of-sin-frac1n-cdot-sin-frac2n-cdot-cdot-sin/406826#406826 – N. S. Jun 04 '13 at 17:50
  • Is it $e^{\int_0 ^1 \ln (\sin x) dx}$? – Hagrid Jun 04 '13 at 18:08
  • @Hagrid: Yes, it is. – Mhenni Benghorbal Jun 04 '13 at 19:04
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    @Hagrid: If you are interested the integral, you can use the technique. – Mhenni Benghorbal Jun 04 '13 at 19:27
  • Thank you. Could you tell me why $\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big),\mathrm{d}x= \frac \pi 2\log(2)+2\int_0^{\pi/2}\log(\sin(x)),\mathrm{d}x $ and how I can use it, if the limits of my integral are $0$ and $1$? – Hagrid Jun 04 '13 at 19:50
  • @Hagrid: because $ \int_0^{\pi/2}\log(\sin(x)),\mathrm{d}x=\int_0^{\pi/2}\log(\cos(x)),\mathrm{d}x .$ – Mhenni Benghorbal Jun 05 '13 at 01:17
  • I know that, but while proving this, we use the fact that $ \int _0 ^{\pi /4} \ln \cos y dy = \int _{\pi / 4} ^{\pi /2} \ln \sin y dy$. Why is it true for $\int_0^1 \ln (\cos x)dx$? – Hagrid Jun 05 '13 at 06:00

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I am not quite sure, but maybe you do not need integration: let us consider $$ \exp \left( \ln \left(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdots \sin 1\right)\frac{1}{n} \right).$$ Now let us expand the sinus: $$\exp \left( \ln \left( \frac{(n-1)!}{n^{n-1}}+ o\left( \frac{1}{n^{n-1}}\right) \right)\frac{1}{n} \right)=\exp \left(\frac{n-1}{n}\ln\left(\frac{n-1}{en}\right)+o(1) \right) \rightarrow \frac{1}{e}.$$ Here I used the Stirling approximation. I hope I haven't done anything wrong!

Gary
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Kore-N
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  • This is incorrect, see the correct answer here: https://math.stackexchange.com/q/406819 – Gary Jan 10 '23 at 13:43