Integrals of products of fractional parts Let $((x)):=x-\lfloor x \rfloor -1/2$, where $\lfloor x \rfloor $ denotes the greatest integer $\le x$. 
Let $a,b,c,...$ denote arbitrary natural numbers. It is clear that
$$ \int_0^a ((x/a)) dx =0.$$
A little exercise shows that 
$$ \int_0^{ab} ((x/a))((x/b)) dx =\frac{\gcd(a,b)^2}{12}.$$
It appears that 
$$ \int_0^{abc} ((x/a))((x/b))((x/c)) dx =0,$$
but I have not been able to determine a general formula for 
$$ \int_0^{abcd} ((x/a))((x/b))((x/c))((x/d))dx.$$
I would be interested in a general formula for the last integral, and for similar integrals with a larger number of factors. It seems that these integrals, or the corresponding sums, such as
$$ \sum_{k=0}^{ab-1} ((k/a))((k/b)) =  \frac{\gcd(a,b)^2-1}{12},$$
should be in the literature, but I have not been able to find a reference. 
 A: One place where I have seen this come up is in a paper of Lemke Oliver and Soundararajan.  Here such quantities appear naturally in computing moments of some remainder terms, and Proposition 3.1 gives some basic properties.  The sums are very closely related to the integrals (see Proposition 3.2 there).  These are not too hard to establish, and I don't know really anything non-trivial about these integrals.  I think it would be quite interesting to understand even the four fold integrals better. 
A: These quantities are studied under the name Franel Integrals in the literature. If we change notation so that $\psi(x) = x−⌊x⌋−1/2 $ is the saw tooth function, we may define the "fourth order" Franel integral as 
$I(a,b,c,d)=\int_{0}^{1} \psi(ax) \psi(bx) \psi(cx) \psi(dx) dx.$
Expanding $\psi(x)$ in a Fourier series we see that:
$I(a,b,c,d) = \frac{1}{16 \pi^4} \sum_{\substack{s,t,u,v \in Z - \{0\}\\ sa+tb+uc+ue=0}} \frac{1}{stuv} =: \frac{1}{16 \pi^4} L(a,b,c,d)$
There is no explicit general closed form evaluation of this quantity but there are many special cases known, including
$L(1,1,1,1) = \frac{\pi^4}{5},$
$L(a,1,1,1) = \left(\frac{1}{3a} - \frac{2}{15a^3} \right) \pi^4,$
$L(a,a,b,b) = \frac{\pi^4}{9} + \frac{8 (a,b)^4 \zeta(4)}{a^2b^2}.$
