# Euler's Choice and Riemann's Oversight?

Euler's Choice:

When Euler crafted the zeta function, he knew that $\zeta(1)$ diverged, so he made $\zeta(1)$ undefined.

When he crafted the zeta generating function using the Bernoulli numbers, he allowed that function to return $-\frac{1}{2}$ for $\zeta(0)$. I believe he could foresee no consequences in doing this.

Riemann's Oversight:

When Riemann modified the zeta function to handle the complex numbers, he defined $\zeta(1)$ to return ComplexInfinity. This would allow the reciprocal to return a $0$ to match the $0$ returned by $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$ when $s=1$. His oversight was to not fix $\zeta(0)$ to also return Complex Infinity for the same reason as he did for $\zeta(1)$.

My Testing:

I modified Mathematica to tweak $\zeta(0)$:

Unprotect[Zeta];
Zeta[0] := ComplexInfinity;
Protect[Zeta];

Replace the middle line with:

Zeta[0]=.;

to restore Zeta back to the original.

So far, my testing has not broken anything. However, we all know that zeta is found in many functions, so we should do exhaustive testing.

A Conjecture:

I think that the Critical Line being on Re $\frac{1}{2}$ is a consequence of a prime being a factor of exactly $\frac{1}{2}$ of the square-free numbers.

The Question:

What do you think the consequences would be if the definition of $\zeta(0)$ is changed?

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what do you mean by: "a prime being a factor of exactly $1/2$ of the square-free numbers" –  draks ... Apr 8 '12 at 20:19
The count of the square-free numbers with $p_{n}$ as the greatest prime factor is twice the count of those with $p_{n-1}$ as the gpf. Therefor, each prime is a factor of $\frac{1}{2}$ of the square-free numbers. –  Fred Kline Apr 8 '12 at 20:27
The claim that $p_n$ is twice as frequently a gpf of squarefrees as $p_{n-1}$ is not the same as "a prime being a factor of exactly 1/2 of the squarefree numbers." –  anon Apr 8 '12 at 23:40
Your claims about squarefree numbers might be considered true if they are ordered by the number of prime factors; but that is incompatible with the natural ordering of natural numbers by size, which is how people interpret statements about "$1/2$ of" some infinite set. –  Greg Martin Apr 9 '12 at 5:16
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For one thing, $\zeta(s)$ would no longer be continuous, much less analytic, at $s=0$; the fact that it currently is analytic there is extremely strong evidence that the value $-1/2$ is the natural value. (In other words, it's not simply a "definition", any more than the square of 13 is "defined" to be 169 - that is simply the value it turns out to have when one applies math consistently.) For another thing, the functional equation would fail (this is the math that leads to the appropriate definition of $\zeta(s)$ when $\Re(s)\le0$).
I also don't know what you mean by "a prime [is] a factor of exactly $1/2$ of the squarefree numbers". The proportion of squarefree numbers that the prime $p$ divides equals $1/(p+1)$. For example, $2$ divides exactly a third of the squarefree numbers.
Your statement about the functional equation with the Re: $s\le0$ would still hold when we change $-\frac{1}{2}$ to $0$. –  Fred Kline Apr 8 '12 at 21:34
@Rudy: Taken literally the functional equation isn't defined at $s=0$ or $1$ just as $0\times\infty$ isn't defined. But $\zeta(s)$ has a simple pole at $s=1$ so the functional equation has a removable singularity there; imposing continuity means we must accept $\zeta(0)=-1/2$. –  anon Apr 8 '12 at 23:37
Depends which form of the functional equation one uses ($\xi(s)$ is entire even if $\zeta(s)$ isn't), but yes, this is a valid point. –  Greg Martin Apr 9 '12 at 5:13