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

- Partial summation
- 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- Why these two methods are so different? (This is very vague though...)
- 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?

usualsetting in the theory of exponential sums. $\endgroup$