Question about the term $\sum_{ \rho} \frac{X^{\rho}}{\rho}$ in the explicit formula of $\sum_{n \leq X} \Lambda(n) \chi(n)$

Let $\Lambda$ be the von Mangoldt function and $\chi$ a primitive character mod $q$, then we have the explicit formula $$\sum_{n \leq X} \Lambda(n) \chi(n) = \delta_{\chi} X - \sum_{ |Im \ \rho| \leq T} \frac{X^{\rho}}{\rho} + O((\frac{X}{T}+1) \log^2(qXT)),$$ where $\delta_{\chi}$ is $1$ if $\chi$ is the principal character and $0$ otherwise, and $\rho$'s are the non-trivial zeros of the $L$ function $L(s, \chi)$. Let us take $T = X^{a}$ where $0< a < 1$. From this formula we can easily deduce that $$| \sum_{ |Im \ \rho| \leq T} \frac{X^{\rho}}{\rho} | \ll X.$$ I was wondering does the bound still hold if I put the absolute value inside the sum?, i.e. do we have $$\sum_{ |Im \ \rho| \leq T} | \frac{X^{\rho}}{\rho} | \ll X.$$ My guess is that it is true but I was not sure how to see this. Any comments would be appreciated. Thank you very much.

• Is $T$ fixed here? If so, then this is trivial from $|X^\rho|=X^{\mathrm{Re}\,\rho}\leq X$. – Wojowu Dec 21 '17 at 11:44
• If $T$ is not fixed, then this is false, because $\sum_{\rho} |\rho|^{-1}$ diverges; see Exercise 10.2.1.1(c) of Montgomery and Vaughan. – Peter Humphries Dec 21 '17 at 12:07
• $\sum_\rho \Re( \frac{x^\rho}{\rho})$ converges because from the functional equation and the density of zeros, there is a sequence $v_n \to \infty$ such that $\frac{\zeta'}{\zeta} (\sigma+iT_n) = \mathcal{O}(\log T_n), \sigma \in [-1,2]$, therefore $\lim_{n \to \infty} \int_{2-iT_n\to 2+iT_n \to -\infty+iT_n \to - \infty - i T_n \to 2-iT_n} \frac{\zeta'}{\zeta}(s) \frac{x^s}{s} ds = \int_{2-i\infty }^{2+i\infty} \frac{\zeta'}{\zeta}(s) \frac{x^s}{s} ds$ and we can apply the residue theorem – reuns Dec 21 '17 at 14:18
• @PeterHumphries $T$ changes but with respect to $X$ so I was wondering if it can still be bounded by $\ll X$. I fixed the question. Thank you. – Johnny T. Dec 21 '17 at 18:08
• I think you can use zero-density estimates to get what you want. See chapters 10,18 of Iwaniec-Kowalski. – Matt Young Dec 21 '17 at 18:49