Prove $\int_a^b \left| \sum_{m \in \mathcal{M}}e(-mx)\right| dx\leq2(b-a+1)\int_0^{\frac{1}{2}}\left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx$

On page 205 of the book 'Analytic Number Theory' by Iwaniec and Kowalski,they give $$\int_a^b \left| \sum_{m \in \mathcal{M}}e(-mx)\right| dx\leq2(b-a+1)\int_0^{\frac{1}{2}}\left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx$$

where $$\mathcal{M}$$ stands for the interval $$[\alpha,\beta]$$ with $$\alpha\, ,\beta \in \mathbb{R}$$, and $$e(x)=e^{2\pi ix}$$.

I wonder how to get this upper bound,but I have tried to give another : $$\int_a^b \left| \sum_{m \in \mathcal{M}}e(-mx)\right| dx \leq \sum _{n=\left[ a\right]}^\left[ b\right] \int_{n}^{n+1}\left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx \\ =([b]-[a]+1)\int_{0}^{1}\left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx \\ \leq(b-a+2)\int_{0}^{1}\left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx$$ Thanks for any help.

• what is the question? Iwaniec-Kowalski do a half interval bound with a factor of $2$, while yours is a full interval with a factor of $1$, so not sure what you want May 25, 2020 at 19:18
• Also, what are $\alpha, \beta$? It would also be nice to define your notation $e(x)$ though I assume it stands for $e^{i2\pi x}$.
– lcv
May 26, 2020 at 3:15

We have \begin{align*} \int_a^b \left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx &= \int_a^{a+\lfloor b-a\rfloor} \left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx + \int_{a+\lfloor b-a\rfloor}^b \left| \sum_{m \in \mathcal{M}}e(-mx)\right|dx\\ &\leq (\lfloor b-a \rfloor + 1) \int_{-1/2}^{1/2} \left|\sum_{m \in \mathcal{M}}e(-mx)\right|dx\\ &\leq 2(b - a + 1) \int_0^{1/2} \left|\sum_{m \in \mathcal{M}}e(-mx)\right|dx \end{align*} since $$\left|\sum_{m \in \mathcal{M}}e(-mx)\right| = \left|\sum_{m \in \mathcal{M}}e(mx)\right|.$$