Integration of Power Series
Summary
Suppose the power series $\sum\limits_{n = 0}^{\infty} c_{n}x^{n}$ converges at $\left| x \right| \lt R$. We then define the function $f$ as follows:
$$ f(x) = \sum\limits_{n = 0}^{\infty} c_{n}x^{n} \qquad \left| x \right| \lt R \tag{1} $$
Then the function $f$ is integrable at $(-R, R)$, and its indefinite integral is as follows:
$$ \int f(x) dx = C + \sum\limits_{n = 0}^{\infty} \dfrac{c_{n}}{n + 1} x^{n+1} \qquad \left| x \right| \lt R \tag{2} $$
Furthermore, the radius of convergence of $f$ and $\displaystyle \int f$ is the same.
Explanation
$(2)$ yields a result as if one integrates the infinite terms of $(1)$ term by term. In other words, one can treat the differentiation of the power series as if polynomial functions are being integrated.
$$ \int \left[ \sum\limits_{n = 0}^{\infty} c_{n}x^{n} \right] dx = \sum\limits_{n = 0}^{\infty} \int c_{n}x^{n} dx $$
It is important to note that the radius of convergence of $\displaystyle \int f$ is the same as that of $f$. This means that the interval of convergence is not necessarily the same, and the convergence at the endpoints of the interval may differ.
Proof
Uniform Convergence and Integrability
Let the sequence of integrable functions $\left\{ f_{n} : f_{n} \text{ is integrable on } [a, b] \right\}$ uniformly converge to $f$ on the interval $[a, b]$. Then $f$ is also integrable on $[a, b]$ and the following holds.
$$ \int_{a}^{b} f(x) dx = \int_{a}^{b} \lim\limits_{n \to \infty} f_{n} (x) dx = \lim\limits_{n \to \infty} \int_{a}^{b} f_{n} (x) dx $$
Suppose $f_{N}(x) = \sum\limits_{n = 0}^{N} c_{n}x^{n}$. Then $f_{N}$ uniformly converges to $f$ on $[a, b] \subset (-R, R)$. By the above lemma, the following holds. At $ \left| x \right| \lt R$,
$$ \begin{align*} \int f(x) dx = \int \lim\limits_{n \to \infty} \sum\limits_{n = 0}^{N} c_{n}x^{n} dx &= \lim\limits_{N \to \infty} \int \sum\limits_{n = 0}^{N} c_{n}x^{n} dx \\ &= C + \lim\limits_{n \to \infty} \sum\limits_{n = 0}^{N} \dfrac{c_{n}}{n + 1} x^{n+1} \\ &= C + \sum\limits_{n = 0}^{\infty} \dfrac{c_{n}}{n + 1} x^{n+1} \end{align*} $$
Additionally, since $\lim\limits_{n \to \infty}\sqrt[n]{\dfrac{1}{n}} = 1$,
$$ \limsup\limits_{n \to \infty} \sqrt[n]{\dfrac{|c_{n}|}{n}} = \limsup\limits_{n \to \infty} \sqrt[n]{|c_{n}|} $$
Therefore, the radius of convergence of the series $\sum\limits_{n = 0}^{\infty} \dfrac{c_{n}}{n + 1} (x-a)^{n+1}$ is the same as that of $f$.
■