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How should I estimate the following integral $$I = \int_0^1 \left( \sum_{n=0}^{p-1} e(n^2t) \right)^2 dt $$ where $p$ is a prime?

Here is the method I followed: \begin{align*} I & = \int_0^1 \left( 1+ \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\ & = 1 + 2 \sum_{n=1}^{p-1} \left( \int_0^1 e(n^2t) dt \right) + \int_0^1 \left( \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\ & = 1 + 2 \sum_{n=1}^{p-1} \left( \frac{e(n^2)-1}{2\pi i n^2} \right) + \int_0^1 \left( \sum_{n=1}^{p-1} e(n^2t) \right)^2 dt \\ & = 1+O((p-1)^{1+\epsilon}) \end{align*} where the third term is estimated using Hua's lemma.

This estimate is not good enough for my purpose. Is it possible to get a better error term than the one here?

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The integral is equal to exactly one - just expand out the square and use orthogonality to get

$$ I = \sum_{0\leq n,m<p} \int_0^1 e((n^2+m^2)t) \mathrm{d} t = \sum_{0\leq n,m<p}1_{n^2+m^2=0}=1.$$

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  • $\begingroup$ Ah, of course. Thanks! Don't know why I overcomplicated it. $\endgroup$ – Iguana May 27 at 14:27

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