From the first part we got to

\(\pi.s\frac{\cos \pi.s}{\sin \pi.s}=1+\displaystyle\sum_{k=1}^{\infty} \frac{1}{(1-\frac{s^2}{k^2})}.-\frac{2s^2}{k^2} \)

Note that \(\frac{\cos \pi.s}{\sin \pi.s}= cotan (\pi.s)\).

We can take the -2 up front and use the geometric series formula to obtain:

\(\displaystyle\pi.s.cotan(\pi.s)=\)

\(=1-2.\displaystyle\sum_{k=1}^{\infty} [1+\frac{s^2}{k^2}+\frac{s^4}{k^4}...].\frac{s^2}{k^2}=1-2.\sum_{k=1}^{\infty} (\sum_{n=1}^{\infty} (\frac{s^2}{k^2})^n)=\)

\(=1-2.\displaystyle\sum_{n=1}^{\infty} \frac{s^{2n}}{1^{2n}}+\frac{s^{2n}}{2^{2n}}...=1-2.\sum_{n=1}^{\infty} \zeta (2n).s^{2n}\)

Now let's try to find a different representation of \(\pi.s.cotan(\pi.s)\)

\(\pi.s.cotan(\pi.s)=\pi.s.\frac{ \cos \pi.s}{ \sin \pi.s}=\)

\(=\displaystyle\pi.s.\frac{e^{i.\pi.s}+e^{-i.\pi.s}}{2}.\frac{2i}{e^{i.\pi.s}-e^{-i.\pi.s}}=\)

\(=\displaystyle\pi.s.i.\frac{e^{i.\pi.s}+e^{-i.\pi.s}}{e^{i.\pi.s}-e^{-i.\pi.s}}\).

Multiplying the numerator and denominator by \(e^{i.\pi.s}\) we get

\(\displaystyle \pi.s.cotan(\pi.s)=\pi.s.i.\frac{e^{2i.\pi.s}+1}{e^{2i.\pi.s}-1}=\)

\(=\displaystyle i.\pi.s + \frac{2.i.\pi.s}{e^{2i.\pi.s}-1}\)

We now need to find an infinite series for \(\frac{z}{e^z-1}\)

Let's assume we have an infinite series representation of that

\(\displaystyle\frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n\)

Then, multiplying both sides by \(e^z-1\), using the Taylor expansion for \(e^z\) and dividing by z, we get

\(1=\displaystyle\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n.\sum_{n=0}^{\infty} \frac{z^{n}}{(n+1)!}\)

Tune in next time for Part 3

## Comments

There are no comments in this discussion.