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I saw a integral like this and try to solve it by using $$ \sin{\frac{\pi}{n}}\sin{\frac{2\pi}{n}}\cdots\sin{\frac{(n-1)\pi}{n}}\ = \frac{n}{2^{n-1}} $$

The $\log$ is based on $e$.

The following is what I try: $$ \int^{\pi}_0 \log{(\sin{x})}dx=\underset{n\rightarrow \infty}{\lim}\overset{n}{\underset{i=1}{\sum}}(\frac{\pi}{n})\log{(\sin{\frac{i\pi}{n}})}\\=\underset{n\rightarrow \infty}{\lim}\frac{\pi}{n}\log{(\sin{\frac{\pi}{n}}\sin{\frac{2\pi}{n}}\cdots\sin{\frac{(n-1)\pi}{n}})}\\=\underset{n\rightarrow \infty}{\lim}\frac{\pi}{n}\log{(\frac{n}{2^{n-1}})}=\pi\underset{n\rightarrow \infty}{\lim}\frac{\log{(\frac{n}{2^{n-1}}})}{n}\\=-\pi\underset{n\rightarrow \infty}{\lim}{\frac{2^{n-1}}{n}\frac{2^{n-1}-n2^{n-1}\log2}{2^{2n-2}}}=-\pi\underset{n\rightarrow \infty}{\lim}{\frac{1-n2^{-1}\log2}{n}}\\=\frac{\log2}{2}\pi $$

But I'm not sure whether I did right for the whole evaluation.

Please help me to check, appreciate.

BTW, my typesetting is by tablet, perhaps it's not neat.

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    $\begingroup$ $\lim_{n\to\infty}\frac{1}{n}\log(\frac{n}{2^{n-1}})=\lim_{n\to\infty}\log(\frac{n}{2^{n-1}})^{\frac{1}{n}}=\log\frac{1}{2}$? $\endgroup$ Commented Feb 5, 2023 at 6:18
  • $\begingroup$ I'm a bit confused by the notation you have used, although I can say that the final answer should be twice that and negative. Edit : (I assume you mean base-10 log by log) $\endgroup$ Commented Feb 5, 2023 at 6:21
  • $\begingroup$ By log, do you mean log base 10 or the natural log? $\endgroup$ Commented Feb 5, 2023 at 6:24
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    $\begingroup$ @xfireskyx yes, after an easy checking, L'Hospital rule is right! so your mistake appears at $$\dots=-\pi\frac{1-n2^{-1}\log2}{n}$$, may be you lost $2^{n-1}$ when multiply. $\endgroup$ Commented Feb 5, 2023 at 7:14
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    $\begingroup$ You are wellcome, best wishes! $\endgroup$ Commented Feb 5, 2023 at 7:18

2 Answers 2

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While the method provided in the other answer is certainly the best way to solve this problem, if you are keen on understanding how to solve it via the limit approach you gave, the following would work:

$$\pi\lim_{n\to\infty}\frac{\ln\left(\frac{n}{2^{n-1}}\right)}{n}$$

From here, we can move the n in the dominator to the exponent of the argument of the logarithm like this:

$$\pi\lim_{n\to\infty}\ln\left(\left(\frac{n}{2^{n-1}}\right)^{1/n}\right)$$

In this case, since the natural logarithm is a continuous function we can swap the order of the limit and the function, yielding:

$$\pi\ln\left(\lim_{n\to\infty}\left(\frac{n}{2^{n-1}}\right)^{1/n}\right)$$

Which simplifies to:

$$\pi\ln\left(\lim_{n\to\infty}\frac{n^{\frac{1}{n}}}{2^{1-\frac{1}{n}}}\right)$$

The limit as $n$ approaches infinity of the denominator is clearly 2 as the exponent simplifies to $1-\frac{1}{\infty}$

The limit of the $n^{th}$ root of $n$ is 1, which can be shown using the following reasoning: $$\lim_{n\to\infty}n^{\frac{1}{n}}=\lim_{n\to\infty}e^{\frac{\ln(n)}{n}}$$ By L'Hopital's Rule:

$$\lim_{n\to\infty}\frac{\ln(n)}{n}=\lim_{n\to\infty}\frac{\frac{1}{n}}{1}=0$$

Therefore: $$\lim_{n\to\infty}n^{\frac{1}{n}}=\lim_{n\to\infty}e^{\frac{\ln(n)}{n}}=e^{0}=1$$

Resulting in $\pi\ln(\frac{1}{2})$ or $-\pi\ln2$

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    $\begingroup$ Nice solution (+1)! $\endgroup$ Commented Feb 5, 2023 at 7:42
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The notation you have used is a bit confusing and your final answer is also wrong. But, this is a very standard question and there is a very neat way to solve this integral, which is as follows $$ \begin{aligned} \int\limits_{0}^{\pi}\ln (\sin(x))dx &= \int\limits_{0}^{\pi/2}\ln(\sin(x))dx+\int\limits_{\pi/2}^{\pi}\ln(\sin(x))dx\\ &= \int\limits_{0}^{\pi/2}\ln(\sin(x))dx+\int\limits_{0}^{\pi/2}\ln(\cos(x))dx\\ &= \int\limits_{0}^{\pi/2}\ln\left(\frac{1}{2}\sin(2x)\right)dx\\ &=\int\limits_{0}^{\pi/2}\ln\left(\sin(2x)\right)dx- \int\limits_{0}^{\pi/2}\ln(2)dx\\ &= \frac{1}{2}\int\limits_{0}^{\pi}\ln(\sin(x))dx-\frac{\pi}{2}\ln 2 \end{aligned} $$

and hence

$$\int\limits_{0}^{\pi}\ln (\sin(x))dx =-\pi\ln 2$$

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  • $\begingroup$ I know that it's right, but I need to think twice what I ignored for the limit, thanks for your help and answer. $\endgroup$
    – xfireskyx
    Commented Feb 5, 2023 at 6:43

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