Proof that $\zeta'(-2n)=(-1)^n\frac{(2n)!}{2(2\pi)^{2n}}\zeta(2n+1)$.

Question

How would one prove the following statement which I found here, and/or does anyone know of a reference with a proof?

$$\zeta'(-2n)=(-1)^n\frac{(2n)!}{2(2\pi)^{2n}}\zeta(2n+1).$$

Answer 1

The proof follows from the reflection formula of $\zeta(s)$ given by $$\zeta(s)=2^{s} \pi^{s-1} \sin \frac{\pi s}{2} \Gamma(1-s)\zeta(1-s).$$

We want to differentiate with respect to $s$ at $s=-2n$. This would seem to be difficult, but note that $\sin\dfrac{\pi s}{2}\Big|_{s=-2n}=-\sin(\pi n)=0$ for integer $n$. Since the derivatives of the rest of the equation turns out to be finite at $s=-2n$, it follows that the only nonzero part of $\zeta'(-2n)$ comes from differentiating the sine function. Thus we have as claimed

\begin{align} \zeta'(-2n) &=\left[2^{s} \pi^{s-1} \Gamma(1-s)\zeta(1-s)\dfrac{d}{ds}\left(\sin\frac{\pi s}{2}\right)\right]_{s=-2n}\\ &=\frac{1}{2}(2\pi)^{-2n} \Gamma(1+2n)\zeta(1+2n)\cos(-\pi n)\\ &=(-1)^n\frac{(2n)!}{2(2\pi)^{2n}}\zeta(2n+1). \end{align}