Let $e(z)$ denote $e^{2 \pi i z}$ and let $f(z)$ a smooth real function. I know one can bound sums of the form $$ \sum_{x \leq X} e(f(x)) $$ via for example Van der Corputs's result, provided we make assumptions on the range of the derivatives. I was wondering if there were results of this type for weighted sums. Can we bound a sum of the form $$ \sum_{x \leq X} g(x) e(f(x)) ? $$ If so I was wondering what kind of assumption would be needed for g(z). (In particular, would it possible for $g(z) = 2^z$, for example? ) Thanks!

2$\begingroup$ if $g(z)=2^z$, then the last summand should give the dominant contribution to the sum. If $g$ varies slowly, then you can use partial summation. Look also in Titchmarsch and in IwaniecKowalski for various results about exponential sums. $\endgroup$– Dorian CoughlanNov 19, 2014 at 17:21
2 Answers
You can use van der Corputs's method for weighted sums as well, see Ch. III in
Karatsuba A. A., Voronin S. M. The Riemann zetafunction Walter de Gruyter & Co., 1992.
What is the exact result you have in mind? I see that a lemma of Van der Corput says something about integrals of functions which oscillate rapidly enough. In the case of a sum I don't see that you could expect the same cancellation to go on.
At any rate, the link has a result similar to what you want using integration by parts. Whatever is true for sums can probably be similarly generalized by a summation by parts .