In the paper https://eudml.org/doc/149759 an estimate for the binary additive divisor problem is given with a power saving. I don't get the main bit of the argument - I'm obviously missing something.
In (75) a power saving (in $n$) is found for the quantity \[ \sum _{t\leq \sqrt {n-1}}\sum _{s\atop {(s,t)=1\atop {s\leq \sqrt {t^2+1}}}}\psi \left (\frac {n}{st}-\frac {\overline s}{t}\right )\] where $\psi $ is the first Bernoulli polynomial (see (14), (39), (40), (73) and the summation conditions on page 174; also take $m=k=1$) and Estermann does this through his "Hilfsatz 7" (page 181). For $t<n^{1/2-\delta }$ we'd get a power saving in $n$ trivially, so I need to understand this Hilfsatz for $t\approx \sqrt n$, so I need to understand how to get a bound \[ \sum _{s=1\atop {(s,t)=1}}^t\psi \left (\frac {c}{s}-\frac {\overline s}{t}\right )\ll n^{1/2-\delta }\] with $c\approx t\approx \sqrt n$. But I don't understand the proof of that Hilfsatz in this case - the quantities $b_j$ in the proof look to all be $\approx 1$ (since $t^{1/4}/c^{1/3}\approx 1/n^{1/24}$) so the equality before (72) looks useless to me (are we not just partitioning the $t$ range into intervals of size $\approx 1$ except for the last one, which has size $t$, on which we therefore cannot apply Hilfsatz 6, since the $s$ range $b-a$ is much too large?).
So the short question is: how does (72) follow from Hilfsatz 6, since (as far as I can see) the $s$ range is too large for Hilfsatz 6 to be applicable? Since this $c\approx \sqrt n$ is the crucial range (I think?), I'm obviously missing something. Can anyone clear it up?
I understand it's a bit difficult to follow my description and to follow the paper in parallel... let me know if I can smooth this in any way. (I think it may help to ignore divisor functions and logs in the paper, since my question only concerns the power savings, and to ignore the $m$ and $k$ since my question concerns in particular the cases $m=k=1$).
