(Topics in Complex Analysis VI) Value of Zeta Function at Even Positive Integers

We come to an interesting applications of residue theorem. In this note, we will evaluate $\zeta(2k)$, where $k$ is a positive integer. Amazingly, we know very little about $\zeta(2k+1)$.

We want to evaluate the infinite sum $\sum_{n=-\infty}^\infty f(n)$, where $f(n)$ is a "good" function, and it has finite poles in the complex plane.  The main idea here is that we combine it with a function $\varphi(z)$, such that poles of $\varphi(x)$ is the whole $\mathbb{Z}$, and they are simple poles. Let $C_N$ be the boundary of the rectangle centered at 0, with midpoints of four edges are $N + \frac{1}{2}, -N - \frac{1}{2}, i(N+\frac{1}{2}), -i(N+\frac{1}{2})$, respectively, where $N$ is a positive integer, s.t, all of poles of $f$ lies inside $C_N$. We then have

$$\int_{C_N}f(z)\varphi(z)dz = \sum_{\substack{n=-N \\ n\neq z_k}}^N Res(f(z)\varphi(z),n) + \sum_{i=1}^n Res(f(z)\varphi(z),z_i)$$

where $z_i$ are poles of $f$. IF the residue of $\varphi(z)$ at each integer point is 1, then

$$\sum_{\substack{n=-N \\ n\neq z_k}}^N Res(f(z)\varphi(z),n) = \sum_{\substack{n=-N \\ n\neq z_k}}^N f(n)$$

we have

$$\int_{C_N}f(z)\varphi(z)dz = \sum_{\substack{n=-N \\ n\neq z_k}}^N f(n) + \sum_{i=1}^n Res(f(z)\varphi(z),z_i)$$

IF $\lim_{N\rightarrow \infty} \int_{C_N}f(z)\varphi(z)dz = 0$, then have

$$\sum_{\substack{n=-\infty \\ n\neq z_k}}^\infty f(n) = -Res(f(z)\varphi(z),z_i)$$

And it is the time for us to compute the residue by familiar limit formula, or the coefficient $c_{-1}$ of the Laurent expansion at pole $z_i$. Now, the problem is what is $\varphi(z)$, and what kind of $f(z)$ is enough good? To choose the simplest $\varphi$ with simple pole at the whole $\mathbb{Z}$, with residue 1, we can think of the function $\varphi(z) = \pi cotg (\pi z)$. It can be checked easily that our assumptions hold. For sufficiently goof $f$, it should decrease fast. In particular, we require $|f(z)|\le \frac{A}{z^2}$, so that $\lim_{z\rightarrow \infty} zf(z) = 0$. We can now prove

Theorem 1. For $f,\varphi$ is described above, the integrand $\lim_{N\rightarrow \infty}\int_{C_N}f(z)\varphi(z)dz = 0$.

Proof. By our familiar inequality $\int_\gamma f \le \max_{z\in\gamma}|f(z)| length(\gamma)$, we have

$$\int_{C_N}f(z)\varphi(z)dz \le \pi (8z + 4) \max_{z\in C_N} |f(z)| |cotg(\pi z)| $$

And it suffices to show that $|cotg(\pi z)|$ is bounded when $z$ runs on the boundary of $C_N$. We have

$$|cotg(\pi z)| = |\frac{cos \pi z}{\sin \pi z}|= |\frac{e^{\pi i z} + e^{-\pi i z}}{e^{\pi i z} - e^{-\pi i z}}| = |\frac{e^{2\pi i z}+1}{e^{2\pi i z} - 1}|$$

When $z$ runs on the edge with $Re(z) = N + \frac{1}{2}$, and $Im(z) = y$, then

$$|\frac{e^{2\pi i z}+1}{e^{2\pi i z} - 1}| = |\frac{1-e^{-2\pi y}}{1 + e^{-2\pi y}}|<1$$

It is very similar when $z$ runs on the edge with $Re(z) = -N - \frac{1}{2}$, and $Im(z) = y$. To prove the same thing for the edge $Re(z) = x$, and $Im(z) = N + \frac{1}{2}$, we use the reverse triangle inequality $|z_1 - z_2| \le ||z_1|-|z_2||$ for the denominator. We have

$$|\frac{e^{2\pi i z}+1}{e^{2\pi i z} - 1}| = \frac{|e^{2\pi i x - \pi(2N+1)}+1|}{|e^{2\pi i x - \pi(2N+1)}-1|}\le \frac{1+e^{-\pi(2N+1)}}{1 - e^{-\pi(2N+1)}} < 2$$

And similarly, we obtain the same bound for the remaining edge. To conclude

$$\int_{C_N}f(z)\varphi(z)dz \le \pi (8z + 4) \max_{z\in C_N} |f(z)| |cotg(\pi z)| < 2\pi(8z+4)f(z)$$

Because $\lim_{z\rightarrow \infty}zf(z) = 0$, we have $\lim_{N\rightarrow\infty}\int_{C_N}f(z)\varphi(z)dz = 0$ (Q.E.D)

It is our time to compute the following sum

$$\sum_{n\ge 1}\frac{1}{n^2}$$

In this case $f(z) = \frac{1}{z^2}$, satisfy our condition for $f$, and note that

$$\sum_{n\ge 1}\frac{1}{n^2} = \frac{1}{2}\sum_{\substack {n = -\infty \\ n\neq0}}^{\infty}\frac{1}{n^2} = \frac{1}{2}S$$

And $f(n)$ has the only pole at $z = 0$. Hence

$$S = -Res(f(z)\pi cotg(\pi z),0) = -Res(\pi \frac{cotg(\pi z)}{z^2},0)$$

Let us represent $cotg (\pi z)$ into Laurent series at 0 to compute the residue, by multiplication of two Laurent series $\cos z$ and $\frac{1}{\sin z}$. We first compute the Laurent expansion of $\frac{1}{\sin \pi z}$ at 0, with note that $\frac{1}{\sin z}$ has the simple pole at $0$, and hence, it has the form

$$\frac{1}{\sin z} = c_{-1}z^-1 + c_0 + c_1z +...$$

To compute $c_i$, we have

$$\sin z \frac{1}{\sin z} = 1 = (z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + ...)(c_{-1}z^-1 + c_0 + c_1z +...)$$

The coefficients for $z^{-1}$ is $0$. The coefficient for $z^0$ is 1 and it is equal $c_{-1}$. Hence, $c^{-1} = 1$. Similarly, the coefficient for $z^1$ is $0$ and equal $c_0$. By this way, we obtain the Laurentz series at 0, which is $\frac{1}{z} + \frac{1}{6}z + \frac{7}{360}z^3 + ... $.

Furthermore, the Laurentz series of $\cos z$ at 0 is $1 - \frac{z^2}{2} + \frac{z^4}{4!} - ...$. Hence, the product of our two series is

$$cotg(z) = (\frac{1}{z} + \frac{1}{6}z + \frac{7}{360}z^3 + ... )(1 - \frac{z^2}{2} + \frac{z^4}{4!} - ...) = \frac{1}{z} - \frac{z}{3} + ...$$

Hence,

$$\pi\frac{cotg(\pi z)}{z^2} = \frac{1}{\pi^3z^3} - \frac{\pi^2}{3z} + ...$$

And the residue $Res(f(z)\pi cotg(\pi z), 0) = -\frac{\pi^2}{3}$. And hence, $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.

What can we see via the evaluation of $\zeta(2)$? Actually, we can evaluate $\zeta(4)$ by the same way, just replace the function $f$. Besides, what we need to care is the coefficient of Laurent series of $cotg(z)$ up to $z^3$. And it is an exercise for you to prove that $\zeta(4) = \frac{\pi^4}{90}$.

It is worth noting that this method cannot be extended to evaluate zeta function at non-negative odd integers, because we cannot similarly evaluate the sums of the form

$$\sum_{\substack{n=-\infty\\n\neq 0}}^\infty f(n)$$

anymore.