assumption on $n(r)$ ensures that the canonical products converge, so that $f$ is a meromorphic function on $\Complex$. Moreover, by assumption $f$ has neither poles nor zeros on the real line.

A computation shows that $E_p'(z)/E_p(z)= -z^p/(1-z)$. Thus $$\label{eq:logderiv}
\frac{f'(x)}{f(x)}= \sum_{n=1}^{\infty}\left( \frac{(x/\overline{a_n})^p}{x-\overline{a_n}}-\frac{(x/a_n)^p}{x-a_n}\right).$$

Let $a\in \Complex$, ${\operatorname{\rm Im}\nolimits}a >0 $, $t\in \Real$. Then $$\begin{aligned}
\label{eq:basic}
 \frac{(t/\overline{a})^p}{t-\overline{a}}-
\frac{(t/a)^p}{t-a} & = \frac{1}{|a|^{2p}}
 t^p \left(\frac{ a^p(t-a)-\overline{a}^p (t-\overline{a})}{|t-a|^2}\right)
\nonumber
\\ & = 2i \frac{1}{|a|^{2p}}
 t^p \left( \frac{ t {\operatorname{\rm Im}\nolimits}(a^p) - {\operatorname{\rm Im}\nolimits}(a^{p+1})}{|t-a|^2}\right).\end{aligned}$$ Now set $a=\alpha + i\beta$, $\beta>0$, and note that $|{\operatorname{\rm Im}\nolimits}a^p| \leq p\beta |a|^{p-1}$. Thus for $x\in \Real$ $$\begin{aligned}
\label{eq:intbd}
\left| \int_0^x \left( \frac{(t/\overline{a})^p}{t-\overline{a}}-
\frac{(t/a)^p}{t-a}\right) dt\right| & 
\leq \frac{2}{|a|^{p+1}}  \left|
\int_0^x \frac{ p|t|^{p+1}\beta + (p+1) |a||t|^p\beta}{(t-\alpha)^2+ \beta^2} dt\right|.\end{aligned}$$ Now for $q>0$ $$\int_0^x  \frac{ t^{q}\beta }{(t-\alpha)^2+ \beta^2} dt
= x^{q} \arctan((x-\alpha)/\beta)- q\int_0^x
t^{q-1} \arctan((t-\alpha)/\beta) dt.$$ Using that for $s\in \Real$, $|\arctan s| <\pi/2$, we find that for $|x|>1$ $$\label{eq:est1}
\left|\int_0^x  \frac{ t^{q}\beta }{(t-\alpha)^2+ \beta^2} dt \right|
\leq C |x|^{q}$$ for some constant $C$, independent of $\alpha$ and $\beta$. To prove the lemma, we will split $\{a_j\}$ into two sets, depending on the relative size of $|a_j|$ and $2|x|$. For $|a_j|\leq 2|x|$, we first note that $$p\int _0^x \frac{ \beta t^{p+1}}{(t-\alpha)^2+\beta^2}dt 
= p\beta \int 