The "seashell constant": closed form for $\frac12\exp\int_0^1-\log(\sin(\frac{\pi}{6}+\frac{2\pi}{3}x))\mathrm dx$?

Question

I am looking for a closed form for $$ R = \frac{1}{2}\,\exp\left(-\int_{0}^{1} \log\left(\sin\left(\frac{\pi}{6} + \frac{2\pi}{3}\,x\right)\right){\rm d}x\right)\approx 0.6159 $$

Wolfram does not give a closed form for $R$.

Wolfram expresses the integral in terms of dilogarithms with complex arguments; is that the simplest way to express $R$ without an integral?

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Where $R$ comes from

On a circular arc with central angle $\dfrac{4\pi}{3}$ and radius $r$, draw $n+1$ equally spaced points, including two at the ends of the arc. Draw line segments from each of these points to the midpoint of the complementary arc.

Here is an example with $n=7$.

Let $P(r,n)=\text{product of lengths of the $n+1$ line segments}$.

INTERESTING FACT: $\lim\limits_{n\to\infty}P(R,n)=R$.

I call this constant "the seashell constant".

Proof

We have

$$P(r,n)=\prod\limits_{k=0}^{n}2r\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}\left(\frac{k}{n}\right)\right)=\exp\left[n\cdot\color{red}{\frac{1}{n}\sum\limits_{k=0}^{n}\log\left(2r\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}\left(\frac{k}{n}\right)\right)\right)}\right]$$

If $P(r,n)$ converges then the red part must converge to $0$.

The red part converges to $\int_0^1\log\left(2r\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}x\right)\right)\mathrm dx$.

Setting this integral equal to $0$ gives $r=\frac12\exp\int_0^1-\log\left(\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}x\right)\right)\mathrm dx$. Call this value $R$.

Now we will show that $\lim\limits_{n\to\infty}P(R,n)=R$.

For ease of notation, let $f(x)=\log\left(\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}x\right)\right)$.

$\begin{align} \lim\limits_{n\to\infty}P(R,n)&=\lim\limits_{n\to\infty}\exp\sum\limits_{k=0}^{n}\left(\log(2R)+f\left(\frac{k}{n}\right)\right)\\ &=\lim\limits_{n\to\infty}\exp\left[\log(2R)+n\log(2R)+\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)+f(0)\right]\\ &=\lim\limits_{n\to\infty}\exp\left[\log(2R)\color{blue}{-n\int_0^1f(x)\mathrm dx+\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)}-\log 2\right]\\ \end{align}$

By considering riemann sums, and noting that $f(0)=f(1)$, we can see that:

$$\lim\limits_{n\to\infty}\left(\color{blue}{-n\int_0^1f(x)\mathrm dx+\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)}\right)=0$$

$\therefore \lim\limits_{n\to\infty}P(R,n)=R$

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I have asked other questions about seashells, for example here and here.

Answer 1

Confirming @KStarGamer's comment:

$\begin{align} R&:=\frac12\exp\int_0^1-\log\left(\sin\left(\frac{\pi}{6}+\frac{2\pi}{3}x\right)\right)\mathrm dx\\ &\overset{x\to\frac{3}{4\pi}x+\frac12}{=}\frac12\exp\left(-\frac{3}{4\pi}\int_{-2\pi/3}^{2\pi/3}\log\left(\cos\left(\frac{x}{2}\right)\right)\mathrm dx\right)\\ &=\frac12\exp\left(-\frac{3}{\color{red}{2}\pi}\int_\color{red}{0}^{2\pi/3}\log\left(\cos\left(\frac{x}{2}\right)\right)\mathrm dx\right)\space\text{because cosine is an even function}\\ &=\frac12\exp\left(\color{red}{\log 2}-\frac{3}{2\pi}\int_0^{2\pi/3}\log\left(\color{red}{2}\cos\left(\frac{x}{2}\right)\right)\mathrm dx\right)\\ &=\frac12\exp\left(\log 2-\frac{3}{2\pi}G\right)\\ &=\boxed{\exp\left(-\frac{3}{2\pi}G\right)}\approx0.615944189614\\ \end{align}$

where $G$ is Gieseking's constant.

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Gieseking's constant seems like a relatively obscure constant (compare MSE searches for Gieseking's constant, Euler-Mascheroni constant, Catalan's constant), but for some reason it has appeared in answers to three seemingly unrelated questions of mine (besides this one, a question about the expected area of a random triangle, and a question about Pascal's triangle).

Answer 2

Using Euler representation of the sine function, the antiderivative does not make any problem and $$I=\int_0^1\log \left(\sin \left(\frac{2 \pi x}{3}+\frac{\pi }{6}\right)\right)\, dx$$ $$I=-\log(2)-i\,\frac {3}{4\pi} \left(\text{Li}_2\left((-1)^{1/3}\right)-\text{Li}_ 2\left(-(-1)^{2/3}\right)\right)$$

But, expanding, this is also $$-\log(2)+\frac{1}{16 \sqrt{3} \pi } \left(\psi ^{(1)}\left(\frac{1}{6}\right)+\psi^{(1)}\left(\frac{2}{6}\right)-\psi^{(1)}\left(\frac{4}{6}\right)- \psi^{(1)}\left(\frac{5}{6}\right)\right)$$

Here we can find the particular value of the trigamma function in terms of $\pi$, Catalan constant and Clausen function.

Formula $(14)$, which is quite recent, is very useful for the computation of $\psi ^{(1)}\left(\frac{p}{q}\right)$. Have also a look at this paper.

I shall not repeat what other users wrote since you have now the result.