Computing zeta at even numbers

Last year I wrote several posts about computing ζ(3) where ζ is the Riemann zeta function. For example, this post.

It happens that ζ can be evaluated in closed form at positive even arguments, but there’s still a lot of mystery about zeta at positive odd arguments.

There’s a way to derive ζ(2n) using contour integration. But you don’t need to be clever about constructing contours if you start from the following result which is derived using contour integration.

Suppose f is analytic except for poles at zk and that the sum

\sum_{n \ne z_k} f(n)

converges, where the sum is over all integer n except poles of f. Then the sum can be computed from residues at the poles:

 \sum_{n \ne z_k} f(n) = -\sum_{k} \text{Res}\left( f(z) \pi \cot \pi z; z_k \right)

Here Res stands for “residue.” The residue of a function g(z) at z0 is defined as the coefficient c−1 of (zz0)−1 in the Laurent series for g. Equivalently,

 c_{-1} = \lim_{z\to z_0} (z - z_0) g(z)

This means we can evaluate infinite sums by taking a limit that may be a simple application of L’Hôpital’s rule.

The theorem above could be used to calculate a lot more than values of the Riemann zeta function by choosing various functions f, but for our purposes f(z) will be z−2m for some positive integer m.

\zeta(2m) = \sum_{n=1}^\infty \frac{1}{n^{2m}} = \frac{1}{2} \sum_{\stackrel{n=-\infty}{n\ne 0} }^\infty \frac{1}{n^{2m}} = -\frac{1}{2}\,\text{Res}\left( \frac{\pi \cot \pi z}{z^{2m}}; 0 \right )

Why can’t this same trick be used for evaluating the Riemann zeta function at odd integer arguments? The second equality above fails. For even powers of n, the sum over the positive integers is half the sum over all integers. For odd powers, the sum over all integers is zero.

If we write out the Laurent series for π cot πz then we can read off the residues of π cot πz/z2m.

\pi \cot \pi z = \frac{1}{z}-\frac{\pi ^2 z}{3}-\frac{\pi ^4 z^3}{45}-\frac{2 \pi ^6 z^5}{945} - \cdots

When we divide by z² the coefficient of z becomes the coefficient of z−1, i.e. the residue. This residue is -π²/3, and so negative one half of this is π²/6, i.e. ζ(2) = π²/6.

The same line of reasoning shows that ζ(4) = π4/90 and ζ(6) = π6/945.

Finding the values of ζ(2m) in general is no harder than finding the power series for cotangent, which is not trivial but not insurmountable either. See this post.

The final result is that

\zeta(2m) = (-1)^{m-1} 2^{2m-1} \pi^{2m} \frac{B_{2m}}{(2m)!}