Non-trivial zeros play an important (main) role in the distribution of prime numbers. Are there theorems in which trivial zeros play an important (main) role?

Since you tagged this 'reference request', I'm going to answer with a theorem in my own paper *"Euler, the symmetric group, and the Riemann zeta function."*

Let $\pi$ be a permutation in the symmetric group $S_n$. An *ascent* is an occurrence of $\pi(j)<\pi(j+1)$ for $1\le j\le n-1$. For
example, the permutation $(24513)$ has $3$ ascents. The *Eulerian number* $\genfrac{\langle}{\rangle}{0pt}{1}{n}{k}$ is defined to be
the number of permutations in $S_n$ with exactly $k$ ascents. (The Eulerian numbers are not to be confused with the
Euler numbers $E_n$.) There is a surprising identity for alternating sums of Eulerian numbers: For integer $n\ge 1$ we have
$$
\zeta(-n)=\frac{ \sum_{k=1}^n (-1)^{k} \genfrac{\langle}{\rangle}{0pt}{0}{n}{k}}{2^{n+1}(1-2^{n+1})}.
$$

Of course, $\zeta(-n)$ can be expressed in closed form in terms of the Bernoulli numbers by $$ \zeta(-n)=-\frac{B_{n+1}}{n+1}, $$ so the theorem is also an identity relating Eulerian numbers to Bernoulli numbers. However, the proof is direct. It's also not not deep; it consists of identities in Concrete Mathematics by Graham, Knuth and Patashnik, along with Abel summation.

A conjecture of Quillen-Lichtenbaum

$$ \lim_{s \to n} (n-s)^{-\mu_n} \zeta_F (-s) = \pm \frac{\mid{K_{2n} (O_F)_{tor} }\mid}{\mid{K_{2n+1} (O_F)_{tor}}\mid} R_{F,n} * 2^{?}\\ $$

where $F$ is the number field like $\mathbb{Q}$ and $O_F$ is the integers therein generalizing $\mathbb{Z}$ in the case of Riemann zeta. $\mu_n$ is the multiplicity of the zero there. That makes sure you don't just get $0$. $K_\bullet (O_F)_{tor} $ means the torsion subgroup of algebraic K theory of that ring. $R_{F,n}$ is a so-called regulator and $2^?$ is for an unknown power of $2$. So you see the left hand side is leading information about when you have zeroes at negative integers.

This is proven in a bunch of cases via Voevodsky. In the $O_F=\mathbb{Z}$ case you just get the numerators and denominators of the Bernoulli numbers. That recovers the $-\frac{B_{n+1}}{n+1}$ above up to signs, powers of 2 and regulators.

So if you are interested in computing anything in the RHS, zooming in on the zeroes of the associated $\zeta_F$ are the most important.

exactformulas (such as Explicit Formulas) they must be included, but the size of their contribution is much less than the critical-strip zeros (since they are far to the left). $\endgroup$ – paul garrett Jul 22 '17 at 19:45