The factor $\frac12$ in the Riemann $\xi$ function:

$$\xi(s)=\frac12 s(s-1)\,\pi^{-s/2}\,\Gamma(s/2)\,\zeta(s)$$

was introduced by Riemann, however appears to be redundant. Once he had arrived at:


the remaining poles at $s=0,1$ could have been removed by the simpler factor $s\,(s-1)$. This would have also made the function entire and retains the reflection formula $\xi(s) = \xi(1-s)$.

A quite plausible explanation as to why the additional factor $\frac12$ was introduced, I found here. It boils down to the following trick:


that serves as the first step towards deriving the well known Fourier integral expression for $\xi(\frac12+it)$. Splitting the LHS into $\Gamma$'s with constant weights, only works when the factor $\frac12$ is introduced and apparently this was a trick known to Riemann.

However, this is not the approach towards the Fourier integral that is attributed to Riemann in the well-known books about the Zeta-function of Titchmarsh (1986-edition p254/255) and Edwards (1974-edition p16/17, p41). These predominantly apply integration by parts to an integral representation of $\xi(s)$ and use a special relation involving $\psi(x)$ and $\psi'(x)$. But this approach doesn't seem to require the factor $\frac12$.

The only reference about the factor $\frac12$ that I have ever come across was in a footnote in a book or paper that (paraphrased) said: 'the factor $\frac12$ was introduced by Riemann in his 1859 paper and has since then stuck' (unfortunately I can't recall the precise source of this quote...).

Q1: Is there any reference to literature about why the factor $\frac12$ was introduced?

Q2: Does it actually matter that the factor $\frac12$ got 'stuck' in $\xi(s)$? It is obviously just a factor, but I could imagine that such a redundant factor would make certain formula less 'beautiful', e.g without the factor, $\xi(0=\xi(1)=1$ and the Hadamard product would simply be: $\displaystyle \xi(s)=\prod_{\rho} \left(1-\frac{s}{\rho}\right)$.

  • 3
    $\begingroup$ A translation of Riemann's manuscript can be found at: claymath.org/sites/default/files/ezeta.pdf. It appears that $\xi(s)$ is defined there (page 3) without the factor of $1/2$ (or the factor of $s$). $\endgroup$
    – Mark Lewko
    Dec 12, 2019 at 0:19
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    $\begingroup$ I agree that factor of 1/2 is ridiculous and should not be there (e.g., without it $\xi(0) = 1$). In Riemann's paper (e.g., at the link posted by Mark Lewko) you see the reason Riemann had the $1/2$: he used not $\Gamma(s)$ but $\Pi(s) = \Gamma(s+1)$, for which $\Pi(n) = n!$. (Let's not get started about why we nowadays prefer the Gamma-function, whose value at $n$ is $(n-1)!$.) Since Riemann used $\Pi(s)$, on line -5 of page 3 he was led to $\Pi(s/2)(s-1)\pi^{-s/2}\zeta(s)$, which is $(1/2)s(s-1)\Gamma(s/2)\pi^{-s/2}\zeta(s)$, and there's the "extra" $1/2$ when using the $\Gamma$-function. $\endgroup$
    – KConrad
    Dec 12, 2019 at 0:57
  • $\begingroup$ Your final comment about the Hadamard product seems strange. The standard formula is $\xi(s)=e^{A+Bs}\prod_\rho (1-s/\rho)e^{s/\rho}$ with $B=-\gamma/2-1+\log\sqrt{4\pi}$, and you've only changed the $e^A$ to be 1 instead of 1/2 $\endgroup$ Dec 12, 2019 at 4:19
  • 1
    $\begingroup$ Riemann extended zeta to the complex plane, but of course Euler had worked with zeta for real $s$ long before, and there $\sum n^{-s}$ seems more natural than $2\sum n^{-s}$. So wasn't Riemann just trying to be consistent with previous practice? $\endgroup$ Dec 12, 2019 at 10:21
  • $\begingroup$ @KConrad. Thanks, that makes a lot of sense. So, his train of thought could have been in the language of the paper: I have $\Pi(s/2-1)\pi^{-s/2}\zeta(s)$ and to remove the remaining poles at $s=0,1$, I have to multiply by $s(s-1)$ (i.e. remove the factor $1/(s(s-1))$ on line -7 of page 3) to obtain $s(s-1)\Pi(s/2-1)\pi^{-s/2}\zeta(s)$. Then he decided to further simplify by introducing a factor $\frac12$ and obtained: $\Pi(s/2)(s-1)\pi^{-s/2}\zeta(s)$ (line -5 of page 3). P.S. if you would be prepared to post your comment as an answer, I would gladly accept it. $\endgroup$
    – Agno
    Dec 12, 2019 at 12:27

1 Answer 1


This factor of $\frac12$ is ultimately due to the fact that the usual integral representation is written as a Mellin transform, that is an integral over $\mathbb{R}^\times_{>0}$, whereas the "correct" integral is over $\mathbb{R}^\times=\mathrm{GL}_1(\mathbb{R})$.


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