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When we want to estimate an exponential sum $$ \sum_{M<m\le M'}e(f(m)) \quad\text{with}\quad 1\le M\le M'\le 2M \quad\text{and}\quad e(x):=\exp(2\pi ix) $$ where $e(x):=\exp(2\pi ix)$ and the phase function $f\colon [M,2M]\to\mathbb{R}$ is assumed to be, e.g. $R$-times continuously differentiable with sufficiently large $R$, a naive expectation (which is wrong if $f'(x)$ is large) is that the sum is approximated by the integral: $$ \sum_{M<m\le M'}e(f(m))\approx\int_{M}^{M'}e(f(u))du. $$ I'm looking at two methods

  1. Partial summation
  2. Poisson summation formula

to prove such an approximation mainly in the Kuzmin--Landau range $|f'(x)|\le\frac{1}{2}$. I always get confused with and surprised at their difference. Largely speaking, I would like to know what makes the difference and if it is possible to connect them.

Method 1. Partial Summation.

The above approximation can be carried out by partial summation: $$ \sum_{M<m\le M'}e(f(m))=\int_{M}^{M'}e(f(u))du-2\pi i\int_{M}^{M'}f'(u)e(f(u))\rho(u)du+O(1), $$ where $\rho(x)$ is the sawtooth function $\rho(u):=\{x\}-\frac{1}{2}$. Without employing the Fourier expansion of $\rho(u)$ (which is almost the Poisson summation formula: the insertion of which gives the Poisson summation formula itself), what we can do on the error is (perhaps) only $$ 2\pi i\int_{M}^{M'}f'(u)e(f(u))\rho(u)du \ll M\sup_{M<u\le M'}|f'(u)|. $$ This bound is intuitively clear since the variance of $e(f(x))$ over $x\in[m,m+1]$ is roughly $\ll|f'(x)e(f(x))|$ and so it accumulates to $M\sup_{M<u\le M'}|f'(u)|$ after summed over $m$. In order to make this bound $O(1)$, we need to assume \begin{equation} \tag{1} f'(x)\ll\frac{1}{M}\quad\text{for $M\le x\le M'$} \end{equation} A slight variant of this normal partial summation is used to approximate $f(x)$ by Taylor expansion in the Weyl--Littlewood--Vinogradov type estimate and also in the Bombieri--Iwaniec/Huxley--Watt type estimate. For example, such usages are in the form $$ \sum_{M<m\le M+N}e(f(m)) =\sum_{0<n\le N}e(P(n))+O(TM^{-r}N^{r}), $$ where $(M,M+N]$ is a suitably chosen subinterval of $(M,M']$, $P(n)$ is the degree $r-1$ Taylor polynomial of $f(x)$ at $x=M$ and we assumed $f^{(r)}(x)\asymp TM^{-r}$. Although the right-hand side is not an integral, the main idea here seems the same as this Method 1 for me.

Method 2. Poisson Summation

We now assume only \begin{equation} \tag{2} |f'(x)|\le\frac{1}{2}\quad\text{for $M\le x\le M'$}, \end{equation} which is much weaker than (1). Then, in the Poisson summation formula $$ \sum_{M<m\le M'}e(f(m)) = \int_{M}^{M'}e(f(u))du +\sum_{h\in\mathbb{Z}\setminus\{0\}}\int_{M}^{M'}e(f(u)-hu)du+O(1), $$ the higher frequency part $|h|\ge1$ is ignorable by the first derivative estimate and so $$ \sum_{M<m\le M'}e(f(m)) = \int_{M}^{M'}e(f(u))du+O(1). $$ Once we applied the Poisson summation formula, this method is also intuitively clear. This is a well-known technique which can be used to prove a certain form of Kuzmin--Landau inequality.

Question

As we have seen, two methods requires the conditions (1) and (2) which are quite different in strength. My question is thus
  1. Why these two methods are so different? (This is very vague though...)
  2. From Method 1 view of point, Method 2 seems rather a magic. Is there any way to carry out Method 2 without using Fourier analysis or only using elementary/direct approach? Is there any way to connect Method 1 and Method 2 without expanding $\rho(u)$ into its Fourier series?

Addendum

Method 2 above indeed has a flaw (which can be a key to the answer?): we need to assume $f'(x)$ is monotonic (or assuming $f''(x)$ is of the prescribed (non-zero) size.)
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    $\begingroup$ What is $e$? What are the assumptions on $f$? $\endgroup$ Commented Jan 3, 2023 at 8:10
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    $\begingroup$ I added a notation for $e(x)$ and an assumption of $f$. Well, for the function $f$, I am just supposing some usual setting in the theory of exponential sums. $\endgroup$
    – snufkin26
    Commented Jan 3, 2023 at 8:37

1 Answer 1

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The connection between these summation formulas has been explored in The Summation Formulae of Euler–Maclaurin, Abel–Plana, Poisson.

The Euler-MacLaurin formula is the "partial summation" formula in the OP, for the Abel-Plana formula see the wiki. The cited paper proves the equivalence of the three formulations. The equivalence of the Euler-MacLaurin and Poisson summation formulas is presented in section 5. I think it qualifies as the "elementary/direct approach" requested by the OP.

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    $\begingroup$ Well, I think the implication "EMSF ==> PSF" in Section 5 of the cited paper had been well-known long before the cited paper. The same proof is already in M. N. Huxley, “Area, Lattice Points and Exponential Sums”, (Clarendon Press, 1996) as Lemma 5.4.2 and the same idea is already in E. C. Titchmarsh, “The theory of the Riemann zeta-Function”, 2nd. ed., (Clarendon Press, 1986) as Lemma 4.7. (I think that it is also in the 1st ed.) $\endgroup$
    – snufkin26
    Commented Jan 3, 2023 at 17:26
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    $\begingroup$ Indeed, what I've meant by "(which is almost the Poisson summation formula: the insertion of which gives the Poisson summation formula itself)" is exactly that proof. (And I'm coming up with this question while reading Huxley's book.) $\endgroup$
    – snufkin26
    Commented Jan 3, 2023 at 17:27

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