MHB The Multiple Sine function

[DreamWeaver]

Please excuse the usual rambling preamble (or should that be pre-ramble?), but last year, when idly searching on-line, I happened to chance upon a truly great, great paper by Shin-ya Koyama and Nobushige Kurokawa, concerning the "Multiple Sine function", $$\mathscr{S}_n(x)$$. The (free) paper in question is found here -->>

http://www1.tmtv.ne.jp/~koyama/recentpapers/ei.pdf

The authors have a number of other on-line papers concerning the same subject matter, but I haven't read any of them; as soon as I saw that first one, I thought "this is a function I want to explore on my own", rather than reading more about it - yet. That being the case, I suspect [read as: expect] that everything I post below will be in one of the 'other' papers, but nonetheless, I thought I'd develop a few properties of the Multiple Sine Function and post them here. ----------------------------------------
Multiple Sine function - the definition:
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I'd highly recommend reading the (short but sweet) PDF paper linked to above, but just in case, I'll very briefly skim over the authors' definition of the Multiple Sine function.

Let

$$\mathcal{P}_r(u) = (1-u)\, \text{exp} \left[ u + \frac{u^2}{2} + \cdots + \frac{u^r}{r} \right] = (1-u)\, \text{exp} \left[ \sum_{j=1}^{j=r} \frac{u^j}{j} \right]$$Then the Multiple Sine function is defined by the infinite product:$$\mathscr{S}_r(x) = \exp\left[ \frac{x^{r-1}}{r-1} \right]\, \prod_{n=1}^{\infty} \left[ \mathcal{P}_r\left( \frac{x}{k} \right) \, \mathcal{P}_r\left( -\frac{x}{k} \right)^{(-1)^{r-1}} \right]^{n^{r-1}}$$Lower order examples include the Double Sine function$$\mathscr{S}_2(x) = e^{x}\, \prod_{k=1}^{\infty} \left[ e^{2x} \left( \frac{1-x/k}{1+x/k} \right)^k \right]$$Triple Sine function

$$\mathscr{S}_3(x) = e^{x^2/2}\, \prod_{k=1}^{\infty} \left[ e^{x^2} \left( 1-\frac{x^2}{k^2} \right)^{k^2} \right]$$and Quadruple Sine function

$$\mathscr{S}_4(x) = e^{x^3/3}\, \prod_{k=1}^{\infty} \left[ e^{2k^2x+2x^3/3} \left( \frac{1-x/k}{1+x/k} \right)^{k^3} \right]$$I'll not reproduce it here, as it's elegantly done in the paper linked above, but Shin-ya Koyama and Nobushige Kurokawa demonstrate that$$\frac{ \mathscr{S}_r'(x) }{ \mathscr{S}_r(x) } = \pi x^{r-1}\cot \pi x$$

And, from there, deduce that

$$\int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z) - \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$This integral suggests deep connections between the Multiple Sine function and the Clausen function, Barnes' G-function, Loggamma function, and a good many other 'higher', special functions.

Now that the preliminaries are out of the way, I'll stop quoting others and start adding a few results of my own... brb (Bandit)Questions, comments, feedback, and other charitable donations would be very much appreciated on this thread -->>

http://mathhelpboards.com/commentary-threads-53/commentary-quot-multiple-sine-function-quot-10779.html

Many thanks!

Gethin :D

 

[DreamWeaver]

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Proposition 1.0:
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For $$0<z<1 \in \mathbb{R}\,$$, the Double Sine Function can be expressed - in closed form - in terms of the Clausen function, $$ \text{Cl}_2(2\pi z)$$: $$(1.1)\quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$

It also satisfies the reflection formula:$$(1.2)\quad \mathscr{S}_2(z)\, \mathscr{S}_2(1-z) = 2\sin \pi z$$Where$$\text{Cl}_{2n}(x) = \sum_{k=1}^{\infty} \frac{\sin kx}{k^{2n}}$$

$$\text{Cl}_{2n+1}(x) = \sum_{k=1}^{\infty} \frac{\cos kx}{k^{2n+1}}$$And, in particular$$\text{Cl}_2(x) = -\int_0^x \log\Bigg| 2\sin \frac{t}{2}\Bigg|\, dt$$

$$\text{Cl}_1(x) = \frac{d}{dx} \text{Cl}_2(x) = -\log\Bigg| 2\sin \frac{x}{2}\Bigg|$$NOTE: In the following proof, as well as all subsequent proofs, I will refer to the formula$$\int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z)

- \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$as the Koyama-Kurokawa Formula .[Or the KK-formula for short]Proof:Setting $$n=2\,$$ in the KK-formula gives:$$\pi z\log(\sin \pi z) - \pi \log \mathscr{S}_2(z) = \int_0^{\pi z}\log(\sin x)\, dx = $$$$\int_0^{\pi z}\log(2\sin x)\, dx - \log 2\, \int_0^{\pi z}\, dx = $$$$\int_0^{\pi z}\log(2\sin x)\, dx -\pi z\, \log 2$$Upon setting $$x=y/2\, $$in the logsine integral, we get$$\int_0^{\pi z}\log(2\sin x)\, dx = \frac{1}{2}\, \int_0^{2\pi z}\log\left(2\sin x\right)\, dx = -\frac{1}{2}\text{Cl}_2(2\pi z)$$Hence$$\log \mathscr{S}_2(z) = z\log(2\sin \pi z) + \frac{1}{2\pi}\text{Cl}_2(2\pi z)$$Exponentiating both sides of the above yields (1.1). $$\Box$$To prove the reflection formula, replace $$z\,$$ with $$1-z\,$$ in (1.1) to obtain:$$\mathscr{S}_2(1-z) = \Bigg( 2\sin (\pi-\pi z) \Bigg)^{1-z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi - 2\pi x) }{2\pi} \right]$$The addition formula for the Sine function makes it clear that$$\sin(\pi-\pi z) = \sin \pi z$$and$$\text{Cl}_2(2\pi-2\pi z) = -\text{Cl}_2(2\pi z)$$Hence$$\mathscr{S}_2(1-z) = \frac{2\sin \pi z}{(2\sin \pi z)^z}\, \frac{1}{ \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right] } = \frac{2\sin \pi z}{ \mathscr{S}_2(z) }$$and$$\mathscr{S}_2(z)\, \mathscr{S}_2(1-z) = 2\sin \pi z$$as was to be shown. $$\Box$$
More to follow shortly... (Heidy)

 

[DreamWeaver]

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Proposition 2.0:
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This is stated without a full proof, since it assumes the following classic result of Euler:

$$\sin \pi x = \pi x\, \prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2} \right)
$$

Since

$$\quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$The Double Sine function has the alternative infinite product representation:$$\quad \mathscr{S}_2(z) = (2\pi z)^z\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right] \, \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2} \right)^z$$

More on this thread soon... (Heidy)

 

[DreamWeaver]

Using (1.1):$$(1.1)\quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$The special values $$\text{Cl}_2(\pi/2) = G$$ [Catalan's constant] and $$\text{Cl}_2(\pi) = 0$$, and the extensive list of values for the Sine function found here, on the Wolfram Functions Site:

Sine: Specific values (subsection 03/02)Formula (1.1) yields the following particular values for the Double Sine function:$$\mathscr{S}_2\left( \frac{1}{12} \right) = \left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)^{1/12}\, \exp \left[ \frac{ \text{Cl}_2(\pi/6)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{10} \right) = \left( \frac{\sqrt{5}-1}{2} \right)^{1/10}\, \exp \left[ \frac{ \text{Cl}_2(\pi/5)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{8} \right) = \left( 2 - \sqrt{2} \right)^{1/16}\, \exp\left[ \frac{ \text{Cl}_2(\pi/4)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{6} \right) = \exp \left[ \frac{ \text{Cl}_2(\pi/3)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{5} \right) = \left( \frac{5-\sqrt{5}}{2} \right)^{1/10}\, \exp \left[ \frac{ \text{Cl}_2(2\pi/5)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{4} \right) = 2^{1/8}\, e^{G/2\pi}$$

$$\mathscr{S}_2\left( \frac{1}{3} \right) = 3^{1/6}\, \exp \left[ \frac{ \text{Cl}_2(2\pi/3)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{1}{2} \right) = \sqrt{2} $$Applying the reflection formula - (1.2) - to the above gives the further values:
$$\mathscr{S}_2\left( \frac{11}{12} \right) = \left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)^{11/12}\, \exp \left[ -\frac{ \text{Cl}_2(\pi/6)}{2\pi} \right]$$$$\mathscr{S}_2\left( \frac{9}{10} \right) = \left( \frac{\sqrt{5}-1}{2} \right)^{9/10}\, \exp \left[ -\frac{ \text{Cl}_2(\pi/5)}{2\pi} \right]$$$$\mathscr{S}_2\left( \frac{7}{8} \right) = \left( 2 - \sqrt{2} \right)^{7/16}\, \exp\left[ -\frac{ \text{Cl}_2(\pi/4)}{2\pi} \right] $$

$$\mathscr{S}_2\left( \frac{5}{6} \right) = \exp \left[ -\frac{ \text{Cl}_2(\pi/3)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{4}{5} \right) = \left( \frac{5-\sqrt{5}}{2} \right)^{2/5}\, \exp \left[ -\frac{ \text{Cl}_2(2\pi/5)}{2\pi} \right]$$

$$\mathscr{S}_2\left( \frac{3}{4} \right) = 2^{3/8}\, e^{-G/2\pi}$$

$$\mathscr{S}_2\left( \frac{2}{3} \right) = 3^{1/3}\, \exp \left[ -\frac{ \text{Cl}_2(2\pi/3)}{2\pi} \right]$$

 

[DreamWeaver]

Next, from the Koyama-Kurokawa Formula $$\int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z)

- \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$The Triple Sine function satisfies the relation: $$\int_0^{\pi z} x\log(\sin x)\, dx = \frac{(\pi z)^{2}}{2} \log(\sin \pi z)

- \frac{\pi^{2}}{2}\log \mathscr{S}_3(z)$$-----------------
Proposition 3.0:
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The Triple Sine function has the following closed form:$$\mathscr{S}_3(z) = (2\sin \pi z)^{z^2}\, \exp \Bigg[ \frac{1}{2\pi^2}\Bigg( 2\pi z\, \text{Cl}_2(2\pi z) + \text{Cl}_3(2\pi z) - \zeta(3) \Bigg) \Bigg]$$
Proof: $$\int_0^{\pi z} x\log(\sin x)\, dx = -\frac{(\pi z)^2}{2} \log 2 + \int_0^{\pi z} x\log(2\sin x)\, dx $$Let $$ x \to y/2\, \Rightarrow \, dx = dy/2\, \Rightarrow$$$$-\frac{(\pi z)^2}{2} \log 2 + \frac{1}{4}\, \int_0^{2 \pi z} y\log\left( 2\sin \frac{y}{2} \right) \, dy = $$$$-\frac{(\pi z)^2}{2} \log 2 + \frac{1}{4} \left[ -y\text{Cl}_2(y)\Bigg|_0^{2\pi z} + \int_0^{2 \pi z} \text{Cl}_2(y)\, dy \right] = $$$$-\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \int_0^{2 \pi z} \text{Cl}_2(x)\, dx = $$$$-\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \sum_{k=1}^{\infty} \frac{1}{k^2}\, \int_0^{2 \pi z} \sin kx\, dx = $$$$-\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \sum_{k=1}^{\infty} \frac{1}{k^2}\, \left[ -\frac{1}{k} \cos kx \right]_0^{2\pi z} = $$$$-\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) - \frac{1}{4}\, \text{Cl}_3(2\pi z) +\frac{1}{4}\, \sum_{k=1}^{\infty}\frac{1}{k^3} = $$$$-\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) - \frac{1}{4}\, \text{Cl}_3(2\pi z) +\frac{\zeta(3)}{4} = $$$$\frac{(\pi z)^{2}}{2} \log(\sin \pi z)
- \frac{\pi^{2}}{2}\log \mathscr{S}_3(z)$$Hence$$\log \mathscr{S}_3(z) = $$$$z^2\log(2\sin \pi z)+ \frac{z\, \text{Cl}_2(2\pi z) }{\pi} +\frac{ \text{Cl}_3(2\pi z) }{2\pi^2} - \frac{\zeta(3)}{2\pi^2}$$And$$\mathscr{S}_3(z) = (2\sin \pi z)^{z^2}\, \exp \Bigg[ \frac{1}{2\pi^2}\Bigg( 2\pi z\, \text{Cl}_2(2\pi z) + \text{Cl}_3(2\pi z) - \zeta(3) \Bigg) \Bigg]$$This proves (3.0).

$$\Box$$