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I am reading about Dirichlet polynomials in the book Analytic Number Theory by Iwaniec-Kowalski. During the proof of Theorem 9.1 for any positive real numbers $T, N$ they define a piecewise linear and continuous function $f(t)$ which is an upper bound for the indicator function of $[0,T]$ and vanishes whenever $t>T+N$ or $t<-N$. Then they use the following bound without comment, $$ x > 1 \Rightarrow \left| \int_{t \in \mathbb R}f(t) x^{it }\mathrm d t \right| =O\left( \frac{1}{N (\log x )^2} \right) .$$

Can anyone justify this inequality?

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    $\begingroup$ Integrate by parts twice, to get $(i \log x)^{-2} \int_{\mathbb R} f''(t) x^{it} dt$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 17, 2020 at 23:12
  • $\begingroup$ Thanks! Can I ask what happens with $f''(t)$ when $t$ is near the points of discontinuouity? $\endgroup$
    – Dr. Pi
    Commented Apr 17, 2020 at 23:15
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    $\begingroup$ OK, that was a little bit too fast; but the idea is the same: integrate by parts twice on each interval $(t_{j-1}, t_j)$ where $f$ is linear. The terms with $(i \log x)^{-1} f(t_j) x^{i t_j}$ cancel out, the terms $(i \log x)^{-2} f'(t_j) x^{i t_j}$ are $O(N^{-1} (\log x)^{-2})$, and what remains is an integral involving $f'' = 0$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 17, 2020 at 23:19
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    $\begingroup$ Yes, this is just the matter of continuity. The above calculation is the usual trick for the Fourier transform, re-written in terms of the Mellin transform. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 18, 2020 at 0:31
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    $\begingroup$ "the said authors"??? $\endgroup$
    – Gerry Myerson
    Commented Apr 18, 2020 at 1:57

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