MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Background: In the 1700s, Euler solved the Basel Problem, which was to solve $\sum_{n=1}^\infty\frac{1}{n^2}$ in closed-form. Euler showed that it was equal to $\frac{\pi^2}{6}$ by first expressing $\frac{\sin(x)}{x}$ in Taylor series form, and then writing it as the normalized product of the linear factors given by its roots.

$\frac{\sin(x)}{x} = (1 - \frac{x}{\pi})(1 - \frac{x}{-\pi})(1 - \frac{x}{2\pi})(1 - \frac{x}{-2\pi})...$

When multiplied out and the x^2 terms are collected, they come out to $-\frac{1}{\pi^2}(\sum_{n=1}^∞\frac{1}{n^2})$, which corresponds to x^2's coefficient of -1/6 in the Taylor Series expansion. Thus, the infinite sum comes out to $\frac{\pi^2}{6}$.

My question I have a function with an infinite number of complex zeroes, is continuous and differentiable everywhere, and has no infinities, complex or otherwise, for finite input (I believe this is called a holomorphic function, but I have yet to take a complex analysis class, so I'm not 100% sure). For reference, my function is $(x-1)\zeta(x)$. At x=1, this function is equal to 1, according to Mathematica.

Please note At x = 0, $(x-1)\zeta(x) = \frac{1}{2}$

If z is a non-trivial zero of $\zeta(x)$, so is 1-z

The trivial zeroes of $\zeta(x)$ are the negative even numbers

In the same fashion that Euler described sin(x)/x as the normalized product of its linear factors, can this function also be expressed as the normalized product of its linear factors? Namely, if $z_k$ is the kth non-trivial zero of the Riemann zeta function with "positive" imaginary component, then is this true:

$(x-1)\zeta(x) = (1/2)\prod_{n=1}^\infty(1-\frac{x}{z_n})(1-\frac{x}{1-z_n})(1+\frac{x}{2n})$

If not, why not?

Additional question In general, the only two classes of functions I can think of without any complex zeroes are non-zero constants and exponentials. So, can all functions be expressed as a product of a constant, an exponential, and its normalized linear factors?

If I need to make anything clearer, please let me know. I'm hastily posting this from a school computer and I might not be as clear as I would like.

share|cite|improve this question
Please read FAQ. This site is for research level questions. Which means, in particular, that BEFORE you ask a question on complex analysis here, you are supposed to take at least an undergraduate course on it. – Alexandre Eremenko Jan 10 '13 at 19:46
The summary may be: finish your complex analysis class, it should include certain product representations. – Gerald Edgar Jan 10 '13 at 20:26
@Gerald: which complex analysis classes? I certainly didn't see Weierstrass factorization in my undergraduate complex analysis class. – Yemon Choi Jan 10 '13 at 22:15
@Gabriel: try various textbooks on complex analysis: e.g. Chapter VII, Section 5 of J. B. Conway's Functions of One Complex Variable – Yemon Choi Jan 10 '13 at 22:17
up vote 6 down vote accepted

Hello Gabriel,

I think you should indeed have a look at the theory of (entire) holomorphic/meromorphic functions, since $x\mapsto (x-1)\zeta(x)$ belongs to that class. You more particularly wish to learn about Hadamard or Weierstrass factorization theorems see Wikipedia, as pointed out in the comment by Andres. There you should reasonnably find the answers to your questions.

share|cite|improve this answer
For the specific case of $(s-1)\zeta(s)$, the Hadamard product representation is given in (you can combine it with a representation for $\Gamma(1+s/2)=(s/2)\Gamma(s/2)$ using – Emil Jeřábek Jan 10 '13 at 17:32
Thank you very much! That is indeed what I was looking for, but I didn't know what it was called. I'll take a look at the articles and try and decipher it. From what it looks like, I had the general idea down, but I was missing the pi exponential and some Gamma function stuff. – Gabriel Benamy Jan 21 '13 at 22:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.