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A long time ago, I came across this question. In the third answer, that drew my interest, we have the calculation of

$$\int_0^{\pi}\log (\sin \theta)d\theta$$

by taking logarithms in both sides of the trigonometric formula

$$\prod_{k=1}^{n-1}\sin (k\pi/n)=\frac{n}{2^{n-1}}$$

and using Riemann sums. My main concern is the following. How can we use the theory of Riemann sums if the function (here $\theta \mapsto \log (\sin \theta)$) is unbounded in the interval of integration (in our case $[0,\pi]$)? Do you know what's going on at this point of the aforementioned solution?

Thanks in advance for your valuable help!

Στέλιος
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    Left- or right-hand Riemann sums can converge to an improper integral if the function is monotone. See here – RRL Dec 02 '17 at 15:27
  • @RRL I will study your link. Thanks. If there is any problem, i 'll let you know. :) – Στέλιος Dec 02 '17 at 15:29
  • More generally if the function is not too oscillatory like $x^{-1} \sin x^{-1}$ on $(0,1]$, then sums with well-chosen tags that avoid singularities will converge. – RRL Dec 02 '17 at 15:40

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You are correct about the fact that Riemann-integrability is usually defined for bounded functions, and $\log\sin\theta$ is not bounded over $(0,\pi)$. On the other hand the boundedness assumption can be dropped in some peculiar circumstances. Indeed, the sequence $$ a_n = \frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=-\pi\log 2+\frac{\pi\log(2n)}{n} $$ is convergent, and for any $n>2$ $$ a_n = \int_{0}^{\frac{n-2}{n}\pi}\log\sin\left(\frac{\pi}{n}\left\lceil\frac{n\theta}{\pi}\right\rceil\right)\,d\theta $$ holds. By the dominated convergence theorem, $\lim_{n\to+\infty} a_n$ equals the value of the Lebesgue (or improper-Riemann) integral $\int_{0}^{\pi}\log\sin\theta\,d\theta$.

Jack D'Aurizio
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  • Sorry for the late response, but how do you get the integral representation of the $a_n$'s, since in the interval $[(n-1)\pi/n,\pi]$ the quantity with the floor function gives 1 in the sine, which means that the integrand is not defined then? – Στέλιος Dec 03 '17 at 20:21
  • @SachpazisStelios: you are right, there was a typo, now fixed. – Jack D'Aurizio Dec 03 '17 at 20:26
  • Maybe you wanted to write $(n-1)\pi/n$ in the upper limit of integration. :) – Στέλιος Dec 03 '17 at 20:29