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The Multiple Sine function

DreamWeaver

Well-known member
Sep 16, 2013
337
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.


----------------------------------------
Multiple Sine function - the definition:
----------------------------------------


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

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


and Quadruple Sine function

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


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 (Bandit)


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 :D
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
-----------------
Proposition 1.0:
-----------------


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... (Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
-----------------
Proposition 2.0:
-----------------


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... (Heidy)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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
Sep 16, 2013
337
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)\)


-----------------
Proposition 3.0:
-----------------


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).

\(\displaystyle \Box\)