Exercise 13, § 2

Functions of a Real Variable, VI, p. 25

Let $f$ be an integer $\geq 1$ , $\chi$ a function defined in $\mathbf{Z}$ , with complex values, such that $\chi(x + f) = \chi(x)$ for every $x \in \mathbf{Z}$ . If one sets

\hat{\chi}(y) = \frac{1}{f} \sum_{x=0}^{f-1} \chi(x) e^{-2\pi i xy/f} \quad \text{for } y \in \mathbf{Z},

one has

\chi(x) = \sum_{y=0}^{f-1} \hat{\chi}(y) e^{2\pi i xy/f} \quad \text{for } x \in \mathbf{Z}.

Deduce from Exercise 12 that one has, for every integer $m \geq 2$ ,

\sum_{n \in \mathbf{Z} - \{0\}} \frac{\chi(n)}{n^m} = -\frac{(2\pi i)^m}{m!} \sum_{y=0}^{f-1} \hat{\chi}(y) B_m \left( \frac{y}{f} \right).

For example, one has

1 - \frac{1}{3^3} + \frac{1}{5^3} - \cdots + \frac{(-1)^n}{(2n+1)^3} + \cdots = \frac{\pi^3}{32}.

Solution

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