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.
