This is something that I've known for some time. Its an expansion of Riemann's $\xi$ function based on the traditional representation $$ \xi(s)=\frac{1}{2}\left(1-s(1-s)\int_{1}^{\infty}\frac{\psi(x)}{x}(x^{s/2}+x^{(1-s)/2}) dx\right) $$ or as I prefer, with $x=u^4$, and observing that $$ \begin{align} \frac{s}{2}&=\frac{1}{2}\left(s-\frac{1}{2}\right)+\frac{1}{4}=z(s)+1/4\\ \frac{1-s}{2}&=-\frac{1}{2}\left(s-\frac{1}{2}\right)+\frac{1}{4}=-z(s)+1/4\\ \end{align} $$ putting $w(s)=s^2-s$ one gets then $$ \xi(s)=\frac{1}{2}\left(1+8w\int_{1}^{\infty}\psi(u^4)\left(\frac{u^{4z}+u^{-4z}}{2}\right)du\right) $$ I always liked this suggestive term right at the beginning of the equation, $s(1-s)$, why is it there and then disappears? Riemann right away makes it disappear in its subsequent calculations in order to get the power series representation. So I've tried to find an expansion using it only! After being unsuccessful for some time, it turns out that its a very simple thing. Just look at the expansion of $\cosh(x)$: $$ \cosh(x)= 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+\frac{x^{2n}}{(2n)!}+\cdots $$ the terms are all even powers. So we have the expansion $$ \cosh(4\ln(u)z) =1+\cdots+\frac{(4 \ln(u) z)^{2n}}{(2n)!}+\cdots $$ Now, just look at $z^{2n}$, or a little closer $(z^2)^n$... now with $z=1/2(s-1/2)$ one gets $$ z^2=\frac{1}{4}\left(s-\frac{1}{2}\right)^2=\frac{1}{4}\left(s^2-s+\frac{1}{4}\right)=\frac{1}{4}\left(w+\frac{1}{4}\right) $$ So, now we just have to use the binomial expansion $$ \left[\frac{1}{4}\left(w+\frac{1}{4}\right)\right]^k $$ and collect the terms around $w$. After all that we get the following representation for $\xi(s)$ $$ \xi(s)=\frac{1}{2}\left(1+8\left\{\int_{1}^{\infty}\psi(u^4)c_{0}(x)dx\right\}w^1+\cdots+8\left\{\int_{1}^{\infty}\psi(u^4)c_{n-1}(x)dx\right\}w^n+\cdots \right) $$ where $$ c_{n}(u)=\frac{2^{n}\sqrt{\pi}}{\sqrt{2}}\frac{(\ln(u))^{n+1/2}}{(n+1)!}I_{n-1/2}(\ln(u)) $$ and $I_{a}(u)$ is the modified Bessel function of the first kind.

Now I think that this is the really natural polynomial expansion of the Riemann $\xi$ function.

So I'm asking, is this new?

  • 1
    $\begingroup$ $\xi(1/2+s)=F(s^2)$ then you are just looking at the Taylor series of $F$ at $1/4$. $\endgroup$ – reuns Feb 8 at 10:08
  • $\begingroup$ A.Neves, Can I get your email id to contact you by email? $\endgroup$ – Sourangshu Ghosh Apr 16 at 17:24

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