_0^x t^{p-1}\left( 1+ \frac{ 2\alpha t -(\alpha^2+\beta^2) }
{(t-\alpha)^2+\beta^2}\right)dt$$ and use (\[eq:intbd\]) and (\[eq:est1\]) to get $$\begin{aligned}
\label{eq:smalla}
\sum_{|a_j|\leq 2|x| } \left| 
\int_0^x 
\left( \frac{(t/\overline{a_j})^p}{t-\overline{a_j}}-
\frac{(t/a_j)^p}{t-a_j}\right) dt\right|
&  \leq C  \sum_{|a_j|\leq 2|x|}
\left( |x|^p |a_j|^{-p} +  |x|^{p-1}|a_j|^{-p+1}
\right).\end{aligned}$$ Since $$\label{eq:apowersum}
\sum_{1\leq |a_j|\leq r}|a_j|^{q} = \int_1^r t^{q}dn(t)= r^qn(r)-n(1)-\int_1^r
qt^{q-1} n(t)dt$$ applying our upper bound on $n(r)$ we get from (\[eq:smalla\]) $$\label{eq:asmallbound}
\sum_{|a_j|\leq 2|x| } \left| 
\int_0^x\left( \frac{(t/\overline{a_j})^p}{t-\overline{a_j}}-
\frac{(t/a_j)^p}{t-a_j}\right) dt\right|\leq
C(|x|^{\rho}+1).$$

Now we bound the contribution of the $a_j$ with $|a_j|> 2 |x|$. For this we use (\[eq:basic\]) more directly. Here $$\begin{aligned}
  & 
\sum_{|a_j|>2|x|}
\left| \int_0^x 
\left( \frac{(t/\overline{a_j})^p}{t-\overline{a_j}}-\frac{(t/a_j)^p}{t-a_j}\right) dt \right|   \\ & 
= 2 \sum_{|a_j|>2|x|}
\left| \int_0^x 
 \frac{1}{|a_j|^{2p}}
 t^p \left( \frac{ t {\operatorname{\rm Im}\nolimits}(a_j^p) - {\operatorname{\rm Im}\nolimits}(a_j^{p+1})}{|t-a_j|^2}\right)dt\right|\\
& \leq C  \sum_{|a_j|>2|x|}
\left| \int_0^x \frac{1}{|a_j|^{2p}}
  \left( \frac{ |t|^{p+1} |a_j|^p + |t|^p|a_j|^{p+1})}{|a_j|^2}\right) dt \right|
\\ & \leq C  \sum_{|a_j|>2|x|} (|x|^{p+2} |a_j|^{-p-2}+|x|^{p+1}|a_j|^{-p-1}).\end{aligned}$$ Applying the analog of (\[eq:apowersum\]) and using the upper bound on $n(r)$ we obtain $$\sum_{|a_j|>2|x|} |a_j|^{-q} \leq C|x|^{\rho-q}$$ provided $q>\rho$, giving us $$\sum_{|a_j|>2|x|}
\left| \int_0^x 
\left( \frac{(t/\overline{a_j})^p}{t-\overline{a_j}}-\frac{(t/a_j)^p}{t-a_j}\right) dt \right| \leq C|x|^\rho.$$ Combined with (\[eq:asmallbound\]), this completes the proof of the lemma.

\[p:complex\] Let $f$ be a function analytic in a neighborhood of $
\Omega=\{ z:\: {\operatorname{\rm Im}\nolimits}z \geq 0, \; |z|\geq 1 \}$. 