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.