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 nontrivial 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.
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$\begingroup$ Is $T$ fixed here? If so, then this is trivial from $X^\rho=X^{\mathrm{Re}\,\rho}\leq X$. $\endgroup$ – Wojowu Dec 21 '17 at 11:44

5$\begingroup$ 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. $\endgroup$ – Peter Humphries Dec 21 '17 at 12:07

$\begingroup$ $\sum_\rho \Re( \frac{x^\rho}{\rho})$ 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 $\lim_{n \to \infty} \int_{2iT_n\to 2+iT_n \to \infty+iT_n \to  \infty  i T_n \to 2iT_n} \frac{\zeta'}{\zeta}(s) \frac{x^s}{s} ds = \int_{2i\infty }^{2+i\infty} \frac{\zeta'}{\zeta}(s) \frac{x^s}{s} ds$ and we can apply the residue theorem $\endgroup$ – reuns Dec 21 '17 at 14:18

$\begingroup$ @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. $\endgroup$ – Johnny T. Dec 21 '17 at 18:08

3$\begingroup$ I think you can use zerodensity estimates to get what you want. See chapters 10,18 of IwaniecKowalski. $\endgroup$ – Matt Young Dec 21 '17 at 18:49