Take $x>0$ large, $t\in \mathbb R$, $q\in \mathbb N$ and a non-principal character $\chi $ mod $q$. If you want, take $t\leq x$. How do I bound

\[ \sum _{n\leq x}\frac {\chi (n)}{n^{it}}?\]

My guess was that this is $\ll \sqrt {qt}$, based on thinking about the $q=1$ case, which is the relatively well known statement

\[ \sum _{n\leq x}\frac {1}{n^{it}}=\text {main term }+\mathcal O\left (1\right ).\]

Euler-Maclaurin and Polya-Vinogradov shows the sum in question to be $\ll t\sqrt q$, which is too weak. But in the $q=1$ case EM may be combined with Van der Corput's summation formula to get an $\mathcal O(1)$ bound. If I try to replicate that argument, I get stuck since I don't know how to bound (for $a\in \mathbb N$ with $(a,q)=1$)

\[ \int _x^\infty \frac {e(ua/q)du}{u^{it}}\]

One idea would be with complex analysis: Perron's formula says for some $c>1$ and any $T>0$ the sum is (the error terms really only being "essentially" as small as stated)

$$\int _{c\pm iT}\frac {L_{\chi }(s+it)ds}{s}+\mathcal O\left (x/T\right )$$

where by the Residue Theorem the main term is, for some $0<c'<1$,

$$\int _{c'\pm iT}\frac {L_{\chi }(s+it)x^sds}{s} +\int _{c'+iT}^{c+iT}\frac {L_{\chi }(s+it)x^sds}{s}+(\text { similar integral }).$$

Taking absolute values gives a total error something like $$\ll x/T+\sqrt (T+|t|)$$ which is also too large. But explicitly computing the vertical integral via the functional equation seems to get better bounds, and give my required result.


Instead of using van der Corput, why don't you express the sum as a complex integral?

To simplify matters, consider a smooth sum, i.e., $$S(x,t) = \sum_n f(n/x) n^{-i t},$$ where $f$ is fixed $C^\infty$ function of compact support (though of course much weaker conditions suffice), with $f(x)=1$ for $0\leq x\leq 1/2$, say, so that you are studying essentially the same sum as before. Then $$S(x,t) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^s M f(s) L(s+i t,\chi) ds$$ for $\sigma>1$.

Now displace the integral all the way to the left of the $y$-axis. You'll pick up a pole at $s=0$ (coming from $M f(s)$). Bound $|L(i t)|$ using the functional equation and the bound $|L(1+ i t,\chi)|\ll \log q t$ (valid for $|t|\gg 1$; see, e.g., Montgomery-Vaughan, Thm. 11.4). There is indeed a factor of essentially $\sqrt{q t}$ coming from the functional equation. The dominant term of the bound will thus be $O(\sqrt{q t} \log q t)$.

(If we knew $L(s,\chi)$ has no real zeroes $\sigma$ with $\sigma>1/2$, it would be $O(\sqrt{q t} \log \log q t)$, by the bound in op. cit., exercise 13.2.6.)

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    $\begingroup$ Then do it without a smooth weight, using Perron's formula. (I was avoiding inessential complications, admittedly very minor ones - or non-existent - in this case.) $\endgroup$ Mar 9 '19 at 14:12
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    $\begingroup$ But always ask yourself: do I really need a smooth weight? People often don't. ("Smoothing is given by nature.") $\endgroup$ Mar 9 '19 at 14:13
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    $\begingroup$ Then try using a smooth weight that is a very close approximation to a brutal one (enough that the error incurred in the approximation is smaller than what you want). The decay will have to be slow at first, so make $T$ large enough (are you choosing $T$ optimally, anyhow?). $\endgroup$ Mar 9 '19 at 15:50
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    $\begingroup$ I just carried out a back-of-the-envelope calculation and I'm not seeing $x^{1/4}$. Double-check - if it keeps coming up, tell us why. I think it's purely technical, however. $\endgroup$ Mar 9 '19 at 16:14
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    $\begingroup$ I get a term $x^s$ inside the integral (see my answer), so it would make sense to let $T$ depend on $x$ as well. Also, be careful with the bound $|L(\sigma+it)|\ll \sqrt{q t}$: on the one hand, things might get a tiny bit worse for $\sigma\to 0^+$ (otherwise said: the implied constant could depend on $\sigma$, though nowhere close to disastrously); on the other hand, for $\sigma$ away from $0$, you get a considerably better bound. Use that fact to counterbalance $x^s$. $\endgroup$ Mar 9 '19 at 18:55

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