Denote by $S(c;n,m)$ Kloosterman's sum. Take $X>0$ and take $n,m\in \mathbb Z$ smaller than a small power of $X$ in modulus. It is known that essentially \[ \sum _{c\sim X}\frac {S(c;n,m)}{c}\ll X^{1/6}\hspace {10mm}(1),\] see Theorem 1 of "Ganguly, Sengupta 2012 - Sums of Kloosterman sums over arithmetic progressions". I would like to know what can be said for the sum \[ \sum _{c\sim X}\frac {S(c;n,m)}{c^{1+it}}\hspace {10mm}(2)\] for $t$ something like $\sqrt X$ (or some other small-ish power of $X$).
My main question is "What can I expect in this case? Can I expect a similar bound to hold?".
Here are some (quite useless) thoughts: If I use partial summation I lose a factor $t$, so I don't want to do that. And if I try to look at the proof of (1) in the above paper: First they replace the unweighted sum in (1) by a weighted sum - this is Proposition 8 and the proof surely goes through when we replace $c$ by $c^{1+it}$, leaving us to estimate \[ \sum _{c\sim X}\frac {S(c;n,m)\phi (4\pi \sqrt {nm}/c)}{c^{1+it}}=\frac {1}{(4\pi \sqrt {nm})^{it}}\sum _{c\sim X}\frac {S(c;n,m)\phi (4\pi \sqrt {nm}/c)}{c}\left (\frac {4\pi \sqrt {nm}}{c}\right )^{it}\] for $\phi $ a smooth weight, which they then tackle using Theorem 7 - see the top of Page 163. I could then look at $x^{it}\phi (x)$ as a new test function but now the derivatives are larger than the original one by factors $t/x$, which will probably be a problem.
Any pointers would be appreciated. Thanks!
