This question is more of a check/validation of a concept.

Suppose I want to study $$\sum _{n\leq X}a_n$$ (e.g. $a_n=d(n)$, the divisor function). As is well-known it's standard practice to replace the sharp cut-off with a smooth function $$\sum _{n}a_nw_X(n),$$ where for some parameter $Y$ to be chosen according to our problem $w_X(t)=1$ for $t\in (1,X)$ and $w_X(t)=0$ for $t>X+Y$, and where $w_X(t)$ is smooth on the whole of $\mathbb R$ with derivatives $w^{(j)}_X(t)\ll _j1/Y^j$. This may lead to a "dual" sum $$\sum _{n}a_n\hat w_X(n)$$ which is more tractable, where $$\hat w_X(n)\approx \int _0^\infty w_X(t)\underbrace {f(nt)}_{\text {e.g. a Bessel function for $a_n=d(n)$}}dt$$ and where I'm being annoyingly imprecise but hopefully someone can find more sense than I in what I'm trying to say.

Suppose I want to study instead $$\sum _{n\leq X}a_ne(n\beta )$$ where $\beta \in (0,1/2)$. If I put in a weight function first and then try to remove the $e(n\beta )$ by partial summation, then I am doomed to fail (right?) because then I have a sharp cut-off again in the integral. So I have to do partial summation first and then put in the weight function, giving me something like \begin{eqnarray} e(X\beta )\sum _{n}a_nw_X(n)-2\pi i\beta \int _1^Xe(v\beta )\sum _na_nw_v(n)dv\\\ \approx \sum _na_n\int _0^\infty f(nt)\left (e(X\beta )w_X(t)-2\pi i\beta \int _1^Xe(v\beta )w_v(t)dv\right )dt. \end{eqnarray} The term in the brackets is I think something like $$\int _1^Xe(v\beta )\frac {d}{dv}w_v(t)dv$$ and it seems to me that the derivative is $\ll 1/v$ so essentially this term looks like $w_X(t)$. That would mean the problem is no different to the problem without the $e(n\beta )$ factor, which seems suspiciously simple and so almost certainly points to a conceptual misunderstanding on my part. (e.g. I'm not sure if I'm oversimplifying something with the two arguments of $w_X(t)$). Can anyone clarify me on this?

Thanks!