Evaluation of the Multiple Sine Function....

[DreamWeaver]

Part 1:Define the Multiple Sine Function by

\(\displaystyle \mathcal{S}_m(x)=\text{exp}\left(\frac{x^{m-1}}{m-1}\right)
\prod_{k=1}^{\infty}\left(\mathcal{P}_m\left(\frac{x}{k}\right)\mathcal{P}_m\left(-\frac{x}{k}\right)^p\right)^q\)

Where \(\displaystyle p=(-1)^{m-1}\,\) and \(\displaystyle q=k^{m-1}\,\) (these exponents weren't showing up so well in the above, hence the abbreviations). The function \(\displaystyle \mathcal{P}_n(z)\,\) is defined by:

\(\displaystyle \mathcal{P}_n(z)=(1-z)\,\text{exp}\left(z+\frac{z^2}{2}+\frac{z^3}{3}+\, \cdots \, +\frac{z^n}{n}\right)\)The challenge is this: for \(\displaystyle 0\le\theta <\pi\,\) and \(\displaystyle m\ge 2\,\), prove that:
\(\displaystyle \int_0^{\theta}x^{m-2}\log(\sin x)\,dx=\frac{\theta^{m-1}}{(m-1)}\log(\sin \theta)-\frac{\pi^{m-1}}{(m-1)}\log\mathcal{S}_m\left(\frac{\theta}{\pi}\right)\)

 

[jvicens]

First, let's rewrite the multiple sine function in a simpler form:

\mathcal{S}_m(x)=e^{\frac{x^{m-1}}{m-1}}\prod_{k=1}^{\infty}(1-k^{-m}x^m)^{-1}

Now, we can use the definition of the multiple sine function to rewrite the integral as follows:

\int_0^{\theta}x^{m-2}\log(\sin x)\,dx=\int_0^{\theta}x^{m-2}\log\left(\prod_{k=1}^{\infty}(1-k^{-m}x^m)^{-1}\right)\,dx

Using the properties of logarithms, we can rewrite the integral as a sum of integrals:

\int_0^{\theta}x^{m-2}\log\left(\prod_{k=1}^{\infty}(1-k^{-m}x^m)^{-1}\right)\,dx=\sum_{k=1}^{\infty}\int_0^{\theta}x^{m-2}\log\left((1-k^{-m}x^m)^{-1}\right)\,dx

Next, we can use the definition of the multiple sine function again to rewrite the integral inside the sum:

\int_0^{\theta}x^{m-2}\log\left((1-k^{-m}x^m)^{-1}\right)\,dx=\int_0^{\theta}x^{m-2}\left(-\log(1-k^{-m}x^m)\right)\,dx

Using the properties of logarithms once again, we can rewrite this integral as:

\int_0^{\theta}x^{m-2}\left(-\log(1-k^{-m}x^m)\right)\,dx=\int_0^{\theta}\left(\frac{x^{m-2}}{k^{-m}x^m-1}\right)\,dx

Now, we can use the geometric series formula to simplify the integral:

\int_0^{\theta}\left(\frac{x^{m-2}}{k^{-m}x^m-1}\right)\,dx=\int_0^{\theta}\left(\frac{x^{m-2}}{1+k^{-m}x^m