Solving Summation of sin with n^2 - Svensl

[svensl]

Hello,
Can anyone give some hints on how to solve this:

$$\sum_{n=0}^{K-1}\frac{sin(2\pi n^2\Delta)}{n}$$

It's just the n^2 that complicates things. I tried re-writing it as

$$Im\sum_{n=0}^{K-1}\frac{e^{j n^2 x}}{n}$$,

where $$x=2\pi \Delta$$
but I cannot solve this either.

Thanks,
svensl

 

[d_leet]

What is delta? If it is an integer than sin(2*pi*k) for any integer k is equal to 0.

 

[svensl]

Thanks for the reply.

Delta is a number between (0, 1(.
BTW, K will later be taken to infinity if that makes a difference.

 

[AlphaNumeric]

Perhaps some well choosen function which has poles at certain places in the complex plane to give that summation as residues might be useful? Then you can use a contour integral and Jordans lemma to turn that sum into an integral along the Reals somehow?

That's without putting pen to paper so I might be way off.