For a prime $p$ and integers $a,m$, $0<a,m<p$ define the (partial Kloosterman) sum
$$ S\_p(a,m) = \sum\_{0<k<m} \exp\left(\frac{2\pi\mathrm{i}}{p}(a x + x^{-1})\right), $$
where $x^{-1}$ is the multiplicative inverse of $x$ modulo $p$. I know $\Im S\_p(a,p)=0$ and that there is the Weil bound
$$ |\Re S\_p(a,p)| \leq 2\sqrt{p}. $$


Is there a similar bound on $\Im S\_p(a,m)$ that holds for all $m=1,\ldots,p$? I imagine this question might have a known answer in the literature, but I couldn't find anything. Conjectured bounds would also be interesting to me.


A related question: is the Weil bound true for partial sums as well, i.e., is it true that
$$\max\_m |\Re S\_p(a,m)|\leq 2\sqrt{p}? $$


If I pretend that $(x+a x^{-1})/p$ is uniformly distributed in $[0,1]$, then the distribution of real and imaginary parts of each term is the same, but all imaginary terms cancel. So intuitively only half as many imaginary terms contribute to $\max\_m |\Im S\_p(a,m)|$ as there are real terms that altogether contribute to make up $2\sqrt{p}$. So I would expect some bound of the form
$$ \max\_m |\Im S\_p(a,m)| \leq 2\sqrt{p/2}, $$
but I am not really convinced this is right (for example: should it be $2\sqrt{p/2}$ because $p$ is halved or $\sqrt{p}$ because the number of terms is halved but $p$ is the same?).



Bounds on imaginary parts of partial Kloosterman sums?
The partial sums, normalized by $\sqrt{p}$, are unbounded, as one varies $a$ over all invertible classes modulo $p$n and lets $p$ go to infinity. This follows from results in [this paper](https://arxiv.org/abs/1410.7892), which also has more precise information on the distribution of the partial sums, including real and imaginary parts.


(By the way, it is more usual to write $ax+x^{-1}$ in the phase of the Kloosterman sum, instead of $x+ax^{-1}$; this emphasizes that it is, as a function of $a$, the discrete Fourier transform of $e\_p(x^{-1})$).


