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.
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 $$\left|\sum_{m \in \mathcal{M}}e(-mx)\right| = \left|\sum_{m \in \mathcal{M}}e(mx)\right|.$$