A double "power series" Suppose $(a_{n}(y))$ is a sequence of locally integrable functions. Suppose the series 
$$\sum_{n=0}^{\infty}a_{n}(y)x^{n}$$
converges absolutely and uniformly for $|y|<1$ and $|x|<1$. It is clear that we can take integral term by term:
$$\sum_{n=0}^{\infty}\left(\int_{-1/2}^{1/2}a_{n}(y)dy\right)x^{n}.$$
My question is: does the above integral also converges absolutely and uniformly for some $|x|<r$ ($r>0$)?  Or it is possible that the series be divergent for all values of $x$?
 A: Uniform Convergence
For any interval $[\alpha,\beta] \subset (-1,1)$, we have
$$\begin{align}\left|\sum_{k=n+1}^m \left(\int_\alpha^\beta a_k(y) \, dy\right) x^k\right| &= \left|\int_\alpha^\beta\left(\sum_{k=n+1}^m a_k(y) x^k  \right)\, dy\right| \\ &\leqslant \int_\alpha^\beta \left|\sum_{k=n+1}^m a_k(y)x^k\right| \, dy \end{align}$$
Since the series $\sum a_n(y) x^n$ is uniformly convergent for $|x|,|y| < 1$, for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m > n > N$ we have
$$\left|\sum_{k=n+1}^m a_k(y)x^k\right| < \frac{\epsilon}{\beta - \alpha},$$
and, thus,
$$\left|\sum_{k=n+1}^m \left(\int_\alpha^\beta a_k(y) \, dy\right) x^k\right| < \epsilon$$
Therefore, the series $\sum \left(\int_\alpha^\beta a_n(y) \, dy\right)\, x^n$ is uniformly convergent by the Cauchy criterion.
Absolute Convergence
Here we must prove that $\sum \left|\int_\alpha^\beta a_n(y) \right||x|^n$ converges given that $ \sum |a_n(y)||x|^n$ is pointwise convergent.  If the convergence with absolute terms is uniform, then this is straightforward. However, there are series that converge uniformly and absolutely but not uniformly-absolutely.
For another approach note that,
$$\tag{*}\begin{align}\sum_{n =1 }^N \left|\int_\alpha^\beta a_n(y) \, dy \right| \, |x|^n &\leqslant \sum_{n =1 }^N \int_\alpha^\beta |a_n(y)|\, |x|^n  \, dy \\ &=\int_\alpha^\beta \left(\sum_{n =1 }^N  |a_n(y)|\, |x|^n\right)  \, dy  \\ \end{align}$$
Since we have absolute convergence of $\sum_{n \geqslant 1} |a_n(y)| |x|^n = F(x,y)$, by the monotone convergence theorem it follows that the limit of the LHS of (*) as $N \to \infty$  either must be finite, if $y \mapsto F(x,y)$ is integrable for each $x$, or equal to $+\infty$. 
That leaves the question of whether or not $F$ must be integrable.
