# The Multiple Sine function

#### DreamWeaver

##### Well-known member
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", $$\displaystyle \mathscr{S}_n(x)$$. 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.

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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

$$\displaystyle \mathcal{P}_r(u) = (1-u)\, \text{exp} \left[ u + \frac{u^2}{2} + \cdots + \frac{u^r}{r} \right] = (1-u)\, \text{exp} \left[ \sum_{j=1}^{j=r} \frac{u^j}{j} \right]$$

Then the Multiple Sine function is defined by the infinite product:

$$\displaystyle \mathscr{S}_r(x) = \exp\left[ \frac{x^{r-1}}{r-1} \right]\, \prod_{n=1}^{\infty} \left[ \mathcal{P}_r\left( \frac{x}{k} \right) \, \mathcal{P}_r\left( -\frac{x}{k} \right)^{(-1)^{r-1}} \right]^{n^{r-1}}$$

Lower order examples include the Double Sine function

$$\displaystyle \mathscr{S}_2(x) = e^{x}\, \prod_{k=1}^{\infty} \left[ e^{2x} \left( \frac{1-x/k}{1+x/k} \right)^k \right]$$

Triple Sine function

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

$$\displaystyle \mathscr{S}_4(x) = e^{x^3/3}\, \prod_{k=1}^{\infty} \left[ e^{2k^2x+2x^3/3} \left( \frac{1-x/k}{1+x/k} \right)^{k^3} \right]$$

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

$$\displaystyle \frac{ \mathscr{S}_r'(x) }{ \mathscr{S}_r(x) } = \pi x^{r-1}\cot \pi x$$

And, from there, deduce that

$$\displaystyle \int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z) - \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$

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 Questions, comments, feedback, and other charitable donations would be very much appreciated on this thread -->>

http://mathhelpboards.com/commentar...y-quot-multiple-sine-function-quot-10779.html

Many thanks!

Gethin Last edited:

#### DreamWeaver

##### Well-known member
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Proposition 1.0:
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For $$\displaystyle 0<z<1 \in \mathbb{R}\,$$, the Double Sine Function can be expressed - in closed form - in terms of the Clausen function, $$\displaystyle \text{Cl}_2(2\pi z)$$:

$$\displaystyle (1.1)\quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$

It also satisfies the reflection formula:

$$\displaystyle (1.2)\quad \mathscr{S}_2(z)\, \mathscr{S}_2(1-z) = 2\sin \pi z$$

Where

$$\displaystyle \text{Cl}_{2n}(x) = \sum_{k=1}^{\infty} \frac{\sin kx}{k^{2n}}$$

$$\displaystyle \text{Cl}_{2n+1}(x) = \sum_{k=1}^{\infty} \frac{\cos kx}{k^{2n+1}}$$

And, in particular

$$\displaystyle \text{Cl}_2(x) = -\int_0^x \log\Bigg| 2\sin \frac{t}{2}\Bigg|\, dt$$

$$\displaystyle \text{Cl}_1(x) = \frac{d}{dx} \text{Cl}_2(x) = -\log\Bigg| 2\sin \frac{x}{2}\Bigg|$$

NOTE: In the following proof, as well as all subsequent proofs, I will refer to the formula

$$\displaystyle \int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z) - \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$

as the Koyama-Kurokawa Formula .[Or the KK-formula for short]

Proof:

Setting $$\displaystyle n=2\,$$ in the KK-formula gives:

$$\displaystyle \pi z\log(\sin \pi z) - \pi \log \mathscr{S}_2(z) = \int_0^{\pi z}\log(\sin x)\, dx =$$

$$\displaystyle \int_0^{\pi z}\log(2\sin x)\, dx - \log 2\, \int_0^{\pi z}\, dx =$$

$$\displaystyle \int_0^{\pi z}\log(2\sin x)\, dx -\pi z\, \log 2$$

Upon setting $$\displaystyle x=y/2\,$$in the logsine integral, we get

$$\displaystyle \int_0^{\pi z}\log(2\sin x)\, dx = \frac{1}{2}\, \int_0^{2\pi z}\log\left(2\sin x\right)\, dx = -\frac{1}{2}\text{Cl}_2(2\pi z)$$

Hence

$$\displaystyle \log \mathscr{S}_2(z) = z\log(2\sin \pi z) + \frac{1}{2\pi}\text{Cl}_2(2\pi z)$$

Exponentiating both sides of the above yields (1.1). $$\displaystyle \Box$$

To prove the reflection formula, replace $$\displaystyle z\,$$ with $$\displaystyle 1-z\,$$ in (1.1) to obtain:

$$\displaystyle \mathscr{S}_2(1-z) = \Bigg( 2\sin (\pi-\pi z) \Bigg)^{1-z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi - 2\pi x) }{2\pi} \right]$$

The addition formula for the Sine function makes it clear that

$$\displaystyle \sin(\pi-\pi z) = \sin \pi z$$

and

$$\displaystyle \text{Cl}_2(2\pi-2\pi z) = -\text{Cl}_2(2\pi z)$$

Hence

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

and

$$\displaystyle \mathscr{S}_2(z)\, \mathscr{S}_2(1-z) = 2\sin \pi z$$

as was to be shown. $$\displaystyle \Box$$

More to follow shortly... #### DreamWeaver

##### Well-known member
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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:

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

Since

$$\displaystyle \quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$

The Double Sine function has the alternative infinite product representation:

$$\displaystyle \quad \mathscr{S}_2(z) = (2\pi z)^z\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right] \, \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2} \right)^z$$

More on this thread soon... #### DreamWeaver

##### Well-known member
Using (1.1):

$$\displaystyle (1.1)\quad \mathscr{S}_2(z) = (2\sin \pi z)^{z}\, \text{exp} \left[ \frac{ \text{Cl}_2(2\pi z) }{2\pi} \right]$$

The special values $$\displaystyle \text{Cl}_2(\pi/2) = G$$ [Catalan's constant] and $$\displaystyle \text{Cl}_2(\pi) = 0$$, 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:

$$\displaystyle \mathscr{S}_2\left( \frac{1}{12} \right) = \left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)^{1/12}\, \exp \left[ \frac{ \text{Cl}_2(\pi/6)}{2\pi} \right]$$

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

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

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

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

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

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

$$\displaystyle \mathscr{S}_2\left( \frac{1}{2} \right) = \sqrt{2}$$

Applying the reflection formula - (1.2) - to the above gives the further values:

$$\displaystyle \mathscr{S}_2\left( \frac{11}{12} \right) = \left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)^{11/12}\, \exp \left[ -\frac{ \text{Cl}_2(\pi/6)}{2\pi} \right]$$

$$\displaystyle \mathscr{S}_2\left( \frac{9}{10} \right) = \left( \frac{\sqrt{5}-1}{2} \right)^{9/10}\, \exp \left[ -\frac{ \text{Cl}_2(\pi/5)}{2\pi} \right]$$

$$\displaystyle \mathscr{S}_2\left( \frac{7}{8} \right) = \left( 2 - \sqrt{2} \right)^{7/16}\, \exp\left[ -\frac{ \text{Cl}_2(\pi/4)}{2\pi} \right]$$

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

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

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

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

#### DreamWeaver

##### Well-known member
Next, from the Koyama-Kurokawa Formula

$$\displaystyle \int_0^{\pi z} x^{n-2}\log(\sin x)\, dx = \frac{(\pi z)^{n-1}}{(n-1)} \log(\sin \pi z) - \frac{\pi^{n-1}}{(n-1)}\log \mathscr{S}_n(z)$$

The Triple Sine function satisfies the relation:

$$\displaystyle \int_0^{\pi z} x\log(\sin x)\, dx = \frac{(\pi z)^{2}}{2} \log(\sin \pi z) - \frac{\pi^{2}}{2}\log \mathscr{S}_3(z)$$

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Proposition 3.0:
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The Triple Sine function has the following closed form:

$$\displaystyle \mathscr{S}_3(z) = (2\sin \pi z)^{z^2}\, \exp \Bigg[ \frac{1}{2\pi^2}\Bigg( 2\pi z\, \text{Cl}_2(2\pi z) + \text{Cl}_3(2\pi z) - \zeta(3) \Bigg) \Bigg]$$

Proof:

$$\displaystyle \int_0^{\pi z} x\log(\sin x)\, dx = -\frac{(\pi z)^2}{2} \log 2 + \int_0^{\pi z} x\log(2\sin x)\, dx$$

Let $$\displaystyle x \to y/2\, \Rightarrow \, dx = dy/2\, \Rightarrow$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 + \frac{1}{4}\, \int_0^{2 \pi z} y\log\left( 2\sin \frac{y}{2} \right) \, dy =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 + \frac{1}{4} \left[ -y\text{Cl}_2(y)\Bigg|_0^{2\pi z} + \int_0^{2 \pi z} \text{Cl}_2(y)\, dy \right] =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \int_0^{2 \pi z} \text{Cl}_2(x)\, dx =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \sum_{k=1}^{\infty} \frac{1}{k^2}\, \int_0^{2 \pi z} \sin kx\, dx =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) + \frac{1}{4}\, \sum_{k=1}^{\infty} \frac{1}{k^2}\, \left[ -\frac{1}{k} \cos kx \right]_0^{2\pi z} =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) - \frac{1}{4}\, \text{Cl}_3(2\pi z) +\frac{1}{4}\, \sum_{k=1}^{\infty}\frac{1}{k^3} =$$

$$\displaystyle -\frac{(\pi z)^2}{2} \log 2 - \frac{\pi z}{2}\text{Cl}_2(2\pi z) - \frac{1}{4}\, \text{Cl}_3(2\pi z) +\frac{\zeta(3)}{4} =$$

$$\displaystyle \frac{(\pi z)^{2}}{2} \log(\sin \pi z) - \frac{\pi^{2}}{2}\log \mathscr{S}_3(z)$$

Hence

$$\displaystyle \log \mathscr{S}_3(z) =$$

$$\displaystyle z^2\log(2\sin \pi z)+ \frac{z\, \text{Cl}_2(2\pi z) }{\pi} +\frac{ \text{Cl}_3(2\pi z) }{2\pi^2} - \frac{\zeta(3)}{2\pi^2}$$

And

$$\displaystyle \mathscr{S}_3(z) = (2\sin \pi z)^{z^2}\, \exp \Bigg[ \frac{1}{2\pi^2}\Bigg( 2\pi z\, \text{Cl}_2(2\pi z) + \text{Cl}_3(2\pi z) - \zeta(3) \Bigg) \Bigg]$$

This proves (3.0).

$$\displaystyle \Box$$