Evaluating the sum $\sum_{k=1}^\infty (-1)^k / k^2$ via a contour integral

Question

I'm evaluating the sum:

\begin{align*} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} \end{align*}

I expressed the sum via a complex contour integration. But I'm not getting out the right answer.

My method: It employs the fact that the function

\begin{align*} \dfrac{1}{\tan \pi z} \end{align*}

has simple poles at all integers $n$. This allows us to write

\begin{align*} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} = \frac{1}{2 i} \oint_C e^{i \pi z} \dfrac{1}{z^2 \tan \pi z} dz \end{align*}

where the contour $C$ is defined in fig (a).

We have

\begin{align*} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} & = \frac{1}{2} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} + \frac{1}{2} \sum_{k=-1}^{-\infty} (-1)^k\frac{1}{k^2} = \frac{1}{4 i} \oint_{C+C'} e^{i \pi z} \dfrac{1}{z^2 \tan \pi z} dz \end{align*}

where the contour $C'$ is defined in fig (a). We complete the path of integration along semicircles at infinity (see fig (b)) since the integrand vanishes there. Since the resulting enclosed area contains no singularities except at $z=0$, we can shrink this contour down to an infinitesimal circle $C_0$ around the origin (see fig (c)). So that

\begin{align} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} = \frac{1}{4 i} \oint_{C_0} e^{i \pi z} \dfrac{1}{z^2 \tan \pi z} dz \end{align}

Expanding the integrand around $z=0$, we get

\begin{align} e^{i\pi z} \dfrac{1}{z^2 \tan \pi z} = (1 + i \pi z - \frac{1}{2!} \pi^2 z^2 + \cdots ) \dfrac{1}{z^2 (\pi z + \frac{1}{3} \pi^3 z^3 + \cdots)} \end{align} \begin{align} = (1 + i \pi z - \frac{1}{2!} \pi^2 z^2 + \cdots ) \dfrac{1}{z^3 \pi (1 + \frac{1}{3} \pi^2 z^2 + \cdots)} \end{align} \begin{align} = \cdots + (1 - \frac{1}{2!} \pi^2 z^2 + \cdots ) \dfrac{1}{z^3 \pi} (1 - \frac{1}{3} \pi^2 z^2 + \cdots) \end{align} \begin{align} = \cdots - \dfrac{5\pi}{6} \frac{1}{z} + \cdots \end{align}

So that

\begin{align*} \sum_{k=1}^\infty (-1)^k \frac{1}{k^2} = \frac{1}{4 i} \oint_{ C_0} e^{i\pi z} \dfrac{1}{z^2 \tan \pi z} dz = \frac{1}{4 i} (-2 \pi i) (-\dfrac{5\pi}{6}) = \frac{5 \pi^2}{12} \end{align*}

Which is the wrong answer, it should be $- \pi^2 /12$. What have I done wrong?

Answer 1

The issue is that the integral does not vanish along the bottom semicircle as its radius goes to infinity.

This is because the magnitude of $e^{i \pi z}$ grows exponentially as $\Im(z) \to -\infty$.

But notice that $\pi \csc(\pi z)$ also has simple poles at the integers with residue $$\operatorname{Res}\left[\frac{\pi }{\sin (\pi z)}, k \right] = \lim_{z \to k}\frac{\pi}{\pi \cos (\pi z)} = \frac{1}{\cos(\pi k)} = (-1)^{k}.$$

In addition to being periodic in the real direction, the function $\pi \csc(\pi z)$ tends to zero as $\Im(z) \to \pm \infty$.

Therefore, if we replace $\pi e^{i \pi z}\cot(\pi z)$ with $\pi \csc(\pi z)$, the integral will vanish along both semicircles.