IMC 2004 · Day 1 · P6

It is clear that the left hand side is well defined and independent of the order of summation, because we have a sum of the type $\sum n^{-4}$, and the branches of the logarithms do not matter because all branches are taken. It is easy to check that the convergence is locally uniform on $\mathbb{C} \setminus \{0,1\}$; therefore, $f$ is a holomorphic function on the complex plane, except possibly for isolated singularities at 0 and 1. (We omit the detailed estimates here.)

The function $\log$ has its only (simple) zero at $z = 1$, so $f$ has a quadruple pole at $z = 1$.

Now we investigate the behavior near infinity. We have $\operatorname{Re}(\log(z)) = \log|z|$, hence (with $c := \log|z|$) $$\begin{align*} \left| \sum (\log z)^{-4} \right| &\le \sum |\log z|^{-4} = \sum (\log|z| + 2\pi i n)^{-4} + O(1) \\ &= \int_{-\infty}^{\infty} (c + 2\pi i x)^{-4}\,dx + O(1) \\ &= c^{-4} \int_{-\infty}^{\infty} (1 + 2\pi i x/c)^{-4}\,dx + O(1) \\ &= c^{-3} \int_{-\infty}^{\infty} (1 + 2\pi i t)^{-4}\,dt + O(1) \\ &\le \alpha (\log|z|)^{-3} \end{align*}$$ for a universal constant $\alpha$. Therefore, the infinite sum tends to 0 as $|z| \to \infty$. In particular, the isolated singularity at $\infty$ is not essential, but rather has (at least a single) zero at $\infty$.

The remaining singularity is at $z = 0$. It is readily verified that $f(1/z) = f(z)$ (because $\log(1/z) = -\log(z)$); this implies that $f$ has a zero at $z = 0$.

We conclude that the infinite sum is holomorphic on $\mathbb{C}$ with at most one pole and without an essential singularity at $\infty$, so it is a rational function, i.e. we can write $f(z) = P(z)/Q(z)$ for some polynomials $P$ and $Q$ which we may as well assume coprime. This solves the first part.

Since $f$ has a quadruple pole at $z = 1$ and no other poles, we have $Q(z) = (z-1)^4$ up to a constant factor which we can as well set equal to 1, and this determines $P$ uniquely. Since $f(z) \to 0$ as $z \to \infty$, the degree of $P$ is at most 3, and since $P(0) = 0$, it follows that $P(z) = z(az^2 + bz + c)$ for yet undetermined complex constants $a, b, c$.

There are a number of ways to compute the coefficients $a, b, c$, which turn out to be $a = c = 1/6$, $b = 2/3$. Since $f(z) = f(1/z)$, it follows easily that $a = c$. Moreover, the fact $\lim\limits_{z \to 1} (z-1)^4 f(z) = 1$ implies $a + b + c = 1$ (this fact follows from the observation that at $z = 1$, all summands cancel pairwise, except the principal branch which contributes a quadruple pole). Finally, we can calculate $$f(-1) = \pi^{-4} \sum_{n\ \mathrm{odd}} n^{-4} = 2 \pi^{-4} \sum_{n \ge 1\ \mathrm{odd}} n^{-4} = 2 \pi^{-4} \left( \sum_{n \ge 1} n^{-4} - \sum_{n \ge 1\ \mathrm{even}} n^{-4} \right) = \frac{1}{48}.$$ This implies $a - b + c = -1/3$. These three equations easily yield $a, b, c$.

Moreover, the function $f$ satisfies $f(z) + f(-z) = 16 f(z^2)$: this follows because the branches of $\log(z^2) = \log((-z)^2)$ are the numbers $2\log(z)$ and $2\log(-z)$. This observation supplies the two equations $b = 4a$ and $a = c$, which can be used instead of some of the considerations above.

Another way is to compute $g(z) = \sum \frac{1}{(\log z)^2}$ first. In the same way, $g(z) = \frac{dz}{(z-1)^2}$. The unknown coefficient $d$ can be computed from $\lim\limits_{z \to 1} (z-1)^2 g(z) = 1$; it is $d = 1$. Then the exponent 2 in the denominator can be increased by taking derivatives (see Solution 2). Similarly, one can start with exponent 3 directly.

A more straightforward, though tedious way to find the constants is computing the first four terms of the Laurent series of $f$ around $z = 1$. For that branch of the logarithm which vanishes at 1, for all $|w| < 1$ we have $$\log(1+w) = w - \frac{w^2}{2} + \frac{w^3}{3} - \frac{w^4}{4} + O(|w|^5);$$ after some computation, one can obtain $$\frac{1}{\log(1+w)^4} = w^{-4} + 2 w^{-2}$$

+ \frac{7}{6} w^{-2} + \frac{1}{6} w^{-1} + O(1). The remaining branches of logarithm give a bounded function. So $$f(1+w) = w^{-4} + 2 w^{-2}$$

+ \frac{7}{6} w^{-2} + \frac{1}{6} w^{-1} (the remainder vanishes) and $$f(z) = \frac{1 + 2(z-1) + \frac{7}{6}(z-1)^2 + \frac{1}{6}(z-1)^3} {(z-1)^4} = \frac{z (z^2 + 4z + 1)}{6 (z-1)^4}.$$

From the well-known series for the cotangent function, $$\lim_{N \to \infty} \sum_{k=-N}^{N} \frac{1}{w + 2\pi i \cdot k} = \frac{i}{2} \cot \frac{iw}{2}$$ and $$\lim_{N \to \infty} \sum_{k=-N}^{N} \frac{1}{\log z + 2\pi i \cdot k} = \frac{i}{2} \cot \frac{i \log z}{2} = \frac{i}{2} \cdot i\, \frac{e^{2i \cdot \frac{i \log z}{2}} + 1} {e^{2i \cdot \frac{i \log z}{2}} - 1} = \frac{1}{2} + \frac{1}{z - 1}.$$ Taking derivatives we obtain $$\sum \frac{1}{(\log z)^2} = -z \cdot \left( \frac{1}{2} + \frac{1}{z-1} \right)' = \frac{z}{(z-1)^2},$$ $$\sum \frac{1}{(\log z)^3} = -\frac{z}{2} \cdot \left( \frac{z}{(z-1)^2} \right)' = \frac{z(z+1)}{2(z-1)^3}$$ and $$\sum \frac{1}{(\log z)^4} = -\frac{z}{3} \cdot \left( \frac{z(z+1)}{2(z-1)^3} \right)' = \frac{z(z^2 + 4z + 1)}{2(z-1)^4}.$$