[proofplan] Define $f$ on the open interval of convergence as the pointwise sum of the given [power series](/page/Power%20Series). Choose a compact interval inside $(c-R,c+R)$ that contains the integration interval, use the [radius of convergence](/theorems/273) and the [Weierstrass M-test](/theorems/261) to obtain [uniform convergence](/page/Uniform%20Convergence) there, and then pass the limit through the [integral](/page/Integral) by a uniform error estimate. Finally, use the fundamental theorem of calculus to evaluate the integral of each monomial. [/proofplan] [step:Define the summed function on the interval of convergence] Define the function \begin{align*} f: (c-R,c+R) &\to \mathbb{R} \\ x &\mapsto \sum_{n=0}^{\infty} a_n(x-c)^n. \end{align*} Because the power series has [radius of convergence](/theorems/265) $R>0$, the series converges for every $x \in (c-R,c+R)$. Thus $f(\alpha)$, $f(\beta)$, and $f(x)$ for $x$ between $\alpha$ and $\beta$ are defined whenever $\alpha,\beta \in (c-R,c+R)$. [/step] [step:Choose a compact interval where the power series converges uniformly] Let $\alpha,\beta \in (c-R,c+R)$. Since $|\alpha-c|<R$ and $|\beta-c|<R$, choose a real number $r$ such that \begin{align*} \max\{|\alpha-c|,|\beta-c|\}<r<R. \end{align*} For every $x \in [c-r,c+r]$ and every $n \in \mathbb{N}\cup\{0\}$, \begin{align*} |a_n(x-c)^n| \le |a_n|r^n. \end{align*} Since $r<R$, the series \begin{align*} \sum_{n=0}^{\infty}|a_n|r^n \end{align*} converges by the defining property of the [radius of convergence](/theorems/262). Therefore the [Weierstrass M-test](/theorems/264), applied with $M_n:=|a_n|r^n$, gives uniform convergence of \begin{align*} \sum_{n=0}^{\infty}a_n(x-c)^n \end{align*} on $[c-r,c+r]$. Since both $\alpha$ and $\beta$ lie in $[c-r,c+r]$, the interval with endpoints $\alpha$ and $\beta$ is contained in $[c-r,c+r]$, so the convergence is uniform on that integration interval. [guided] The goal of this step is to replace pointwise convergence by uniform convergence on a closed interval containing all points of integration. Since $\alpha,\beta \in (c-R,c+R)$, both endpoint distances from the centre $c$ are strictly smaller than $R$: \begin{align*} |\alpha-c|<R, \qquad |\beta-c|<R. \end{align*} Hence the maximum of these two numbers is also strictly smaller than $R$, and we may choose $r$ with \begin{align*} \max\{|\alpha-c|,|\beta-c|\}<r<R. \end{align*} Now fix $x \in [c-r,c+r]$. Then $|x-c|\le r$, so for every $n \in \mathbb{N}\cup\{0\}$, \begin{align*} |a_n(x-c)^n| = |a_n|\,|x-c|^n \le |a_n|r^n. \end{align*} Define $M_n:=|a_n|r^n$. This bound is independent of $x$, which is the point of choosing the closed interval $[c-r,c+r]$. Because $r<R$, the defining property of the radius of convergence gives convergence of \begin{align*} \sum_{n=0}^{\infty}M_n = \sum_{n=0}^{\infty}|a_n|r^n. \end{align*} The [Weierstrass M-test](/theorems/272) therefore applies to the functions $x \mapsto a_n(x-c)^n$ on $[c-r,c+r]$ and yields uniform convergence of the power series on that interval. Finally, the inequality $\max\{|\alpha-c|,|\beta-c|\}<r$ implies that the whole interval between $\alpha$ and $\beta$ lies in $[c-r,c+r]$, so the same uniform convergence holds on the interval of integration. [/guided] [/step] [step:Pass the limit through the integral using the uniform error bound] For each $N \in \mathbb{N}\cup\{0\}$, define the partial-sum polynomial \begin{align*} S_N: [c-r,c+r] &\to \mathbb{R} \\ x &\mapsto \sum_{n=0}^{N} a_n(x-c)^n. \end{align*} Let $I_{\alpha,\beta}:=[\min\{\alpha,\beta\},\max\{\alpha,\beta\}]$ denote the compact interval with endpoints $\alpha$ and $\beta$. The preceding step gives $S_N \to f$ uniformly on $I_{\alpha,\beta}$. Since each $S_N$ is continuous and a uniform limit of continuous real-valued functions is continuous, $f$ is continuous on $I_{\alpha,\beta}$. Therefore $f-S_N$ is continuous on $I_{\alpha,\beta}$, hence Borel measurable and bounded on $I_{\alpha,\beta}$; because $\mathcal{L}^1(I_{\alpha,\beta})=|\beta-\alpha|<\infty$, it is Lebesgue integrable. Thus \begin{align*} \sup_{x \in I_{\alpha,\beta}}|f(x)-S_N(x)| \to 0 \end{align*} as $N \to \infty$. Using the oriented convention $\int_{\alpha}^{\beta}=-\int_{\beta}^{\alpha}$ when $\alpha>\beta$, the triangle inequality for the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \left| \int_{\alpha}^{\beta} f(x)\,d\mathcal{L}^1(x) - \int_{\alpha}^{\beta} S_N(x)\,d\mathcal{L}^1(x) \right| &= \left| \int_{\alpha}^{\beta} (f(x)-S_N(x))\,d\mathcal{L}^1(x) \right| \\ &\le |\beta-\alpha| \sup_{x \in I_{\alpha,\beta}}|f(x)-S_N(x)|. \end{align*} The right-hand side tends to $0$, so \begin{align*} \int_{\alpha}^{\beta} f(x)\,d\mathcal{L}^1(x) = \lim_{N\to\infty} \int_{\alpha}^{\beta} S_N(x)\,d\mathcal{L}^1(x). \end{align*} By linearity of the integral over finite sums, \begin{align*} \int_{\alpha}^{\beta} S_N(x)\,d\mathcal{L}^1(x) = \sum_{n=0}^{N}a_n \int_{\alpha}^{\beta}(x-c)^n\,d\mathcal{L}^1(x). \end{align*} Therefore \begin{align*} \int_{\alpha}^{\beta} f(x)\,d\mathcal{L}^1(x) = \sum_{n=0}^{\infty}a_n \int_{\alpha}^{\beta}(x-c)^n\,d\mathcal{L}^1(x). \end{align*} [/step] [step:Evaluate each monomial integral and substitute into the series] For each $n \in \mathbb{N}\cup\{0\}$, define \begin{align*} G_n: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto \frac{(x-c)^{n+1}}{n+1}. \end{align*} Then $G_n$ is differentiable and $G_n'(x)=(x-c)^n$ for every $x \in \mathbb{R}$. By the [Fundamental Theorem of Calculus](/theorems/632), \begin{align*} \int_{\alpha}^{\beta}(x-c)^n\,d\mathcal{L}^1(x) = G_n(\beta)-G_n(\alpha) = \frac{(\beta-c)^{n+1}-(\alpha-c)^{n+1}}{n+1}. \end{align*} Substituting this identity into the term-wise integral formula gives \begin{align*} \int_{\alpha}^{\beta} f(x)\,d\mathcal{L}^1(x) = \sum_{n=0}^{\infty}a_n \int_{\alpha}^{\beta}(x-c)^n\,d\mathcal{L}^1(x) = \sum_{n=0}^{\infty} a_n \frac{(\beta-c)^{n+1}-(\alpha-c)^{n+1}}{n+1}. \end{align*} This is the asserted term-wise integration formula. [/step]