# Dedekind sums + strange integral

I need to write $\displaystyle \int_0^1 f(x)\,dx$, where $f(x) = \# \Sigma_{pq} \cap (x, x+1)$ for $x \in [0, 1]$, where $$\Sigma_{pq} = \left \{ \frac{k}{p} + \frac{l}{q}:\ 1 \leq k \leq p-1, 1 \leq l \leq q-1 \right\}$$ ($p, q$ are coprime positive integers), using Dedekind sums: $s(p,q)$ and $s(q,p)$ where $s(p,q) = \sum_{j=0}^{q} \left( \left( \frac{j}{q} \right) \right) \left( \left(\frac{jp}{q} \right) \right)$, $((x)) = x - \lfloor x \rfloor - \frac{1}{2}$ for $x \not\in \mathbb{Z}$ and $0$ for $x \in \mathbb{Z}$.

I don't know what to start with. Any ideas?

The only thing I've got now (but don't know if it helps) is: $$\int_0^1 f(x)\, dx = \sum_{i=1}^{p-1} \sum_{j=1}^{q-1} \int_0^1 \chi_{(x,x+1)}\left(\frac{i}{p} + \frac{j}{q}\right)\,dx$$

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No. First of all, I really want those Dedekind sums to be there. And which sum equals $\frac{(q-1)(p-1)}{2}$? That integral? As far as I know, it doesn't. – user29683 Jun 2 '12 at 19:07
You are correct. I made a mistake with a sign. – Eric Naslund Jun 2 '12 at 19:34
i've found this: 1, 2, 3. don't know yet if it's useful here but maybe... – user29683 Jun 3 '12 at 17:42