TITLE: Generalized quadratic Gauss sums
QUESTION [6 upvotes]: I was wondering whether anyone knows how to approach the following two
generalizations of the quadratic Gauss sum:
Given integers r,s with gcd(r,s)=1 and integers a,b,N
$F(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $
$G(r,s,N,a,b) = \sum_{w = 0}^{rsa}(-1)^{b w}(\sin\frac{\pi w}{r})(\sin\frac{\pi w}{s}) \exp(\pi i w^2\frac{N}{2 rs}) $
Note that removing the sine terms and the sign, setting a = 2, N = 4, r = 1 and s = prime gives the classical quadratic Gauss sum.
Some experimentation suggests that
$F(r,s,N,a,b) = 0$ for all integers b,N, r,s if a is even and (r,s) =1 and
$G(r,s,N,a,b) = 0$ for all a,b,N and r,s with (r,s) =1
Is there a good reason for these sums to vanish? Or a clean proof/reference?
Is it possible to evaluate F in the case a = 1? It seems to be non-zero then.
I tried reducing to the original Gauss sum by completing the square but 
this seems to get quite ugly.
More generally, do such Gauss-like sums have a more natural generalization
that turns up somewhere?
Thanks

REPLY [2 votes]: I think this should just be computations. Define as usual $e(x) = \exp(2 \pi i x)$. Define
$$
 f(r,s,N,a,b) = \sum_{w=1}^{rsa} e\left(\frac{N}{4 rs} w^2 + \left(\frac{1}{2s}+\frac{b}{2} \right) w\right)
$$
Then $F(a,r,s,N,a,b) = \frac{1}{2i } (f(r,s,N,a,b) - f(r,s,N,a, -b))$, at least if I didn't make any computational mistakes. 
Now, start with $a = 1$. Then you can use, the method described in http://en.wikipedia.org/wiki/Quadratic_Gauss_sum#Generalized_quadratic_Gauss_sums . 
Next, one has to understand what happens if one passes from $a$ to $a + 1$. For this compute
$$
 f(r,s,N,a+1,b) - f(r,s,N,a,b)
$$
I guess, the result should be of the form $z f(\hat r,\hat s, \hat  N, 1, \hat b)$ with $|z| = 1$.
Note: I began summing at $1$ on purpose, so one sums over the group $\mathbb{Z}_{rsa}$, which seems like the correct choice ...