How to compute the Riemann zeta function at negative integers?

Question

There already questions such as $1 + 1 + 1 +\cdots = -\frac{1}{2}$ and Why does $1+2+3+\cdots = -\frac{1}{12}$? which show how $\zeta(0)$ and $\zeta(-1)$ can be calculated.

What are some ways to evaluate the Riemann zeta function at any negative integer that appears to have a direct correlation to $\sum_{n=1}^\infty\frac1{n^s}$? That is to say, results that can be attained by manipulating this series somewhat directly. So not things such as the reflection formula or the Bernoulli numbers which don't seem to relate to the above series.

Answer 1

One can relate the Riemann zeta function $\zeta(s)$ to the Dirichlet eta function $\eta(s)$ using

$$ (1-2^{1-s})\zeta(s) =(1-2^{1-s})\sum_{n=1}^\infty\frac{1}{n^s} =\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} =\eta(s)$$

which follows quite easily by multiplying out the LHS. This identity holds a priori for $s > 1$, and then extends to all of $s\in\mathbb{C}$ by the principle of analytic continuation. Moreover,

$$\eta(s) =\lim_{x\to-1^+}\sum_{n=1}^\infty\frac{x^{n-1}}{n^s}. \tag{1}$$

Indeed, Abel's theorem immediately establish this for $s > 0$, where $\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$ holds, and a more delicate argument shows that this identity continues to hold for all of $s\in\mathbb{C}$. Using this, we get

$$\zeta(0) =-\lim_{x\to-1^+}\sum_{n=1}^\infty x^{n-1} =-\lim_{x\to-1^+}\frac{1}{1-x} =-\frac{1}{2}$$

$$\zeta(-1) =-\frac{1}{3}\lim_{x\to-1^+}\sum_{n=1}^{\infty} nx^{n-1} =-\frac{1}{3}\lim_{x\to-1^+}\frac{1}{(1-x)^2} =-\frac{1}{12}$$

Answer 2

$$\zeta(s) =\sum_{n=1}^\infty n^{-s}=\sum_{n=1}^\infty s\int_1^\infty x^{-s}dx=s \int_1^\infty \lfloor x\rfloor x^{-s-1}dx= \frac{s}{s-1}+\frac12-s \int_1^\infty (x-\lfloor x\rfloor-\frac12)x^{-s-1}dx$$

$$B_0(x)=x-\frac12,\qquad {B_{k+1}}'(x)=B_k(x),\qquad\int_1^2 B_{k+1}(x)dx=0$$

By induction starting with $k=0$, for $\Re(s) > -k$ $$\zeta(s)=\frac{s}{s-1}-\sum_{m=0}^k B_m(1)(\prod_{l=0}^m (s+l))- (\prod_{l=0}^k (s+l)) \int_1^\infty B_k(x-\lfloor x\rfloor)x^{-s-1-k}dx$$ (integration by parts) $$ =\frac{s}{s-1}-\sum_{m=0}^{k+1}B_m(1)(\prod_{l=0}^m (s+l))-( \prod_{l=0}^{k+1} (s+l) )\int_1^\infty B_{k+1}(x-\lfloor x\rfloor)x^{-s-1-(k+1)}dx$$

Whence

$$\zeta(-N)=\frac{-N}{-N-1}-\sum_{m=0}^N B_m(1)\prod_{l=0}^m (-N+l)$$

Answer 3

A self-contained approach. One might start with $$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx $$ where the series and the integral are convergent over $\text{Re}(s)>1$. An analytic continuation over a larger subset of $\mathbb{C}$ can be achieved by applying IBP multiple times: for instance $$ \frac{1}{(s-1)\Gamma(s)}\int_{0}^{+\infty}x^{s-1}\cdot-\frac{d}{dx}\left(\frac{x}{e^x-1}\right)\,dx $$ provides an analytic continuation over $\text{Re}(s)>0$ and $$ \frac{(-1)^{k}}{(s-1)\Gamma(s+k+1)}\int_{0}^{+\infty}x^{s+k}\cdot\frac{d^{k+2}}{dx^{k+2}}\left(\frac{x}{e^x-1}\right)\,dx $$ provides an analytic continuation over $\text{Re}(s)>-(k+1)$. In particular, by picking $s=-k$ in the above line, $$ \zeta(-k)=\frac{(-1)^{k+1}}{k+1}\int_{0}^{+\infty}\frac{d^{k+2}}{dx^{k+2}}\left(\frac{x}{e^x-1}\right)\,dx =\frac{(-1)^k}{k+1}\cdot\left.\frac{d^{k+1}}{dx^{k+1}}\left(\frac{x}{e^x-1}\right)\right|_{x=0}$$ from which it follows that the values of the Riemann zeta function over the non-positive integers can be computed from the Maclaurin series of $\frac{x}{e^x-1}$. Since $$ \frac{x}{e^x-1}=-\frac{x}{2}+\underbrace{\frac{x}{2}\coth\frac{x}{2}}_{\text{even function}}$$ we have that $\zeta(-2n)=0$ for any $n\in\mathbb{N}^+$, and $\zeta(0)=-\frac{1}{2}$.