# Why are the trivial zeros of the Riemann zeta function only negative?

The functional equation of the Riemann zeta function is $$\zeta(s)=2^s\pi^{s-1}\sin(s\pi/2)\Gamma(1-s)\zeta(1-s)$$ clearly $2^s$ and $\pi^{1-s}$ are never equal to zero on the complex plane, and neither is Gamma. Some of the zeros can be determined by $\sin(s\pi/2)=0$ but this is the case when $s=2n$ This would imply that there is a zero at every even integer, but it's known that the only non-trivial zeros are at the negative even integers. What am I doing wrong here?

• The zeros at the negative even integers are called trivial zeros, because you can find them easily with the functional equation. Every other zero is called non-trivial. – Paul K Jan 28 '16 at 8:42

The Riemann zeta function satisfies the functional equation (known as the Riemann functional equation or Riemann's functional equation) $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) \!,$$ where Γ(s) is the gamma function, which is an equality of meromorphic functions valid on the whole complex plane. This equation relates values of the Riemann zeta function at the points s and 1 − s. Owing to the zeros of the sine function, the functional equation implies that ζ(s) has a simple zero at each even negative integer $s = −2n$ — these are known as the trivial zeros of ζ(s). When $s$ is an even positive integer, the product $\sin(πs/2)Γ(1−s)$ on the right is regular and non-zero because $Γ(1−s)$ has a simple pole: the functional equation thus relates the values of the Riemann zeta function at odd negative integers and even positive integers.
• Hi, take a look at my 6 line proof using $\Gamma(s)\zeta(s)$ – reuns Sep 11 '17 at 11:14
The Gamma function has poles at the negative integers. These cancel the zeroes from the $\sin$ for positive even integers due to the $1-s$ in the Gamma function.
For $\Re(s) > 1$ $$\Gamma(s) \zeta(s) = \sum_{n=1}^\infty n^{-s} \int_0^\infty y^{s-1}e^{-y}dy = \sum_{n=1}^\infty \int_0^\infty x^{s-1}e^{-nx}dx\\ = \int_0^\infty x^{s-1}\sum_{n=1}^\infty e^{-nx}dx=\int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$ And by definition of the Bernouilli numbers as the Taylor coefficients of $\frac{x}{e^x-1}$ we obtain $$\Gamma(s) \zeta(s)-\sum_{k=0}^K \frac{B_k}{k!} \frac{1}{s+k-1} = \int_0^\infty x^{s-2}\underbrace{(\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!} x^k 1_{x < 1})}_{= \ \mathcal{O}(x^{K+1})}dx$$ which converges and is analytic for $\Re(s) > -K$. Therefore the poles of $\Gamma(s) \zeta(s)$ are at $\mathbb{Z}_{\le 1}$ of order $1$ and residue $\frac{B_{k-1}}{(k-1)!}$.
Finally, $\Gamma(s+1) = s \Gamma(s)$ implies the poles of $\Gamma(s)$ are at $\mathbb{Z}_{\le 0}$ of order $1$ and residue $\frac{(-1)^k}{k!}$,
so that $\zeta(s)$ has a single pole at $s=1$ and some zeros at $2 \mathbb{Z}_{\le -1}$ where $B_{2k+1} = 0$ (which follows from $\frac{x}{e^x-1}+\frac{x}{2}=\frac{x (e^{x/2}+e^{-x/2})}{e^{x/2}-e^{-x/2}}$ being even)