[proofplan] We prove the result on an arbitrary compact interval contained in the interval of convergence. On such an interval, the original [power series](/page/Power%20Series) and its formal derivative series converge uniformly by comparison with an absolutely convergent numerical series. We then pass from polynomial partial sums to the limit using the fundamental identity obtained by integrating the derivatives of the partial sums. Finally, integrating the finite partial sums term by term and passing to the limit gives the stated integral formula. [/proofplan] [step:Establish uniform convergence of the power series and its formal derivative on compact subintervals] Let $K \subset (x_0-R,x_0+R)$ be a compact interval. Define \begin{align*} s := \sup_{x \in K} |x-x_0|. \end{align*} Then $s<R$. Choose $r$ with $s<r<R$. Since $r$ lies inside the [radius of convergence](/theorems/273), the numerical series \begin{align*} \sum_{n=0}^{\infty} |c_n|r^n \end{align*} converges. For every $x \in K$ and every $n \in \mathbb{N}\cup\{0\}$, \begin{align*} |c_n(x-x_0)^n| \le |c_n|s^n \le |c_n|r^n. \end{align*} Therefore the series defining $f$ converges uniformly on $K$ by comparison with the convergent numerical series $\sum_{n=0}^{\infty}|c_n|r^n$. Now define the formal derivative series \begin{align*} g: (x_0-R,x_0+R) &\to \mathbb{R} \\ x &\mapsto \sum_{n=1}^{\infty} n c_n (x-x_0)^{n-1}. \end{align*} For $x \in K$ and $n \ge 1$, \begin{align*} |n c_n(x-x_0)^{n-1}| \le n|c_n|s^{n-1} = |c_n|r^n \frac{n}{r}\left(\frac{s}{r}\right)^{n-1}. \end{align*} Since $0 \le s/r<1$, the sequence \begin{align*} \left(\frac{n}{r}\left(\frac{s}{r}\right)^{n-1}\right)_{n=1}^{\infty} \end{align*} is bounded. Choose $M=M(s,r)>0$ such that \begin{align*} \frac{n}{r}\left(\frac{s}{r}\right)^{n-1} \le M \end{align*} for every $n \ge 1$. Hence \begin{align*} |n c_n(x-x_0)^{n-1}| \le M |c_n|r^n. \end{align*} The series $\sum_{n=1}^{\infty}M|c_n|r^n$ converges, so the formal derivative series converges uniformly on $K$. [guided] The local nature of the theorem is important: differentiability at a point only requires control on a compact interval around that point. Let $K \subset (x_0-R,x_0+R)$ be a compact interval, and define \begin{align*} s := \sup_{x \in K}|x-x_0|. \end{align*} Because $K$ stays strictly inside the open interval of convergence, we have $s<R$. Choose a number $r$ satisfying $s<r<R$. The definition of the [radius of convergence](/theorems/262) gives absolute convergence at $x=x_0+r$, so \begin{align*} \sum_{n=0}^{\infty}|c_n|r^n \end{align*} converges. For $x \in K$, the inequality $|x-x_0|\le s$ gives \begin{align*} |c_n(x-x_0)^n| \le |c_n|s^n \le |c_n|r^n. \end{align*} Thus every term of the power series on $K$ is bounded by the $n$-th term of a convergent numerical series. This proves [uniform convergence](/page/Uniform%20Convergence) of the original power series on $K$. We next handle the formal derivative. Define \begin{align*} g: (x_0-R,x_0+R) &\to \mathbb{R} \\ x &\mapsto \sum_{n=1}^{\infty} n c_n (x-x_0)^{n-1}. \end{align*} For $x \in K$ and $n\ge 1$, \begin{align*} |n c_n(x-x_0)^{n-1}| \le n|c_n|s^{n-1} = |c_n|r^n \frac{n}{r}\left(\frac{s}{r}\right)^{n-1}. \end{align*} The extra factor $n$ is harmless because it is multiplied by a geometric decay factor. Since $s/r<1$, the sequence \begin{align*} \left(\frac{n}{r}\left(\frac{s}{r}\right)^{n-1}\right)_{n=1}^{\infty} \end{align*} is bounded; choose $M=M(s,r)>0$ such that \begin{align*} \frac{n}{r}\left(\frac{s}{r}\right)^{n-1}\le M \end{align*} for every $n\ge 1$. Therefore \begin{align*} |n c_n(x-x_0)^{n-1}| \le M |c_n|r^n. \end{align*} The comparison series $\sum_{n=1}^{\infty}M|c_n|r^n$ converges, so the formal derivative series converges uniformly on $K$. [/guided] [/step] [step:Pass the derivative identity from polynomial partial sums to the limit] Fix $x \in (x_0-R,x_0+R)$. Choose a compact interval $K=[\alpha,\beta]$ such that $x \in K \subset (x_0-R,x_0+R)$. For each integer $N\ge 0$, define the polynomial partial sum \begin{align*} S_N: K &\to \mathbb{R} \\ t &\mapsto \sum_{n=0}^{N} c_n(t-x_0)^n. \end{align*} Then \begin{align*} S_N'(t)=\sum_{n=1}^{N} n c_n(t-x_0)^{n-1} \end{align*} for every $t \in K$. By the previous step, $S_N \to f$ uniformly on $K$ and $S_N' \to g$ uniformly on $K$, where \begin{align*} g(t)=\sum_{n=1}^{\infty} n c_n(t-x_0)^{n-1}. \end{align*} For $y,t \in K$, the [fundamental theorem of calculus](/theorems/632) for the polynomial $S_N$ gives \begin{align*} S_N(t)-S_N(y)=\int_y^t S_N'(u)\,d\mathcal{L}^1(u). \end{align*} Passing to the limit as $N\to\infty$ is justified by uniform convergence of $S_N$ and $S_N'$ on $K$, and yields \begin{align*} f(t)-f(y)=\int_y^t g(u)\,d\mathcal{L}^1(u) \end{align*} for all $y,t\in K$. Since $g$ is a uniform limit of continuous functions on $K$, it is continuous on $K$. Therefore the last identity implies that $f$ is differentiable at every interior point of $K$ and \begin{align*} f'(t)=g(t)=\sum_{n=1}^{\infty} n c_n(t-x_0)^{n-1}. \end{align*} In particular, this holds at the originally fixed point $x$. Since $x$ was arbitrary, the derivative formula holds throughout $(x_0-R,x_0+R)$. [guided] Fix a point $x \in (x_0-R,x_0+R)$. We prove differentiability near $x$ by working on a compact interval around it. Choose a compact interval $K=[\alpha,\beta]$ such that \begin{align*} x \in K \subset (x_0-R,x_0+R). \end{align*} For each integer $N\ge 0$, define the polynomial partial sum \begin{align*} S_N: K &\to \mathbb{R} \\ t &\mapsto \sum_{n=0}^{N} c_n(t-x_0)^n. \end{align*} Because $S_N$ is a polynomial, it is differentiable on $K$, and its derivative is \begin{align*} S_N'(t)=\sum_{n=1}^{N} n c_n(t-x_0)^{n-1}. \end{align*} The previous step gives two uniform convergence statements on $K$: \begin{align*} S_N \to f \end{align*} uniformly, and \begin{align*} S_N' \to g, \end{align*} uniformly, where \begin{align*} g: K &\to \mathbb{R} \\ t &\mapsto \sum_{n=1}^{\infty} n c_n(t-x_0)^{n-1}. \end{align*} The reason we use the derivative partial sums is that each $S_N$ satisfies the exact finite-dimensional identity \begin{align*} S_N(t)-S_N(y)=\int_y^t S_N'(u)\,d\mathcal{L}^1(u) \end{align*} for every $y,t\in K$. This is the fundamental theorem of calculus applied to the polynomial $S_N$. We now pass this identity to the limit. The left-hand side converges to $f(t)-f(y)$ because $S_N \to f$ uniformly on $K$. The right-hand side converges to $\int_y^t g(u)\,d\mathcal{L}^1(u)$ because $S_N' \to g$ uniformly on $K$; indeed, \begin{align*} \left|\int_y^t (S_N'(u)-g(u))\,d\mathcal{L}^1(u)\right| \le |t-y| \sup_{u\in K}|S_N'(u)-g(u)| \to 0. \end{align*} Thus \begin{align*} f(t)-f(y)=\int_y^t g(u)\,d\mathcal{L}^1(u) \end{align*} for all $y,t\in K$. Since each $S_N'$ is continuous and $S_N'\to g$ uniformly on $K$, the function $g$ is continuous on $K$. Therefore the function \begin{align*} t \mapsto \int_y^t g(u)\,d\mathcal{L}^1(u) \end{align*} is differentiable at every interior point of $K$, with derivative $g(t)$. Consequently $f$ is differentiable at every interior point of $K$, and \begin{align*} f'(t)=g(t)=\sum_{n=1}^{\infty} n c_n(t-x_0)^{n-1}. \end{align*} Since the original point $x$ lies in the interior of some such compact interval $K$, the derivative formula holds at $x$. The point $x$ was arbitrary, so the formula holds on the whole interval of convergence. [/guided] [/step] [step:Integrate the polynomial partial sums and pass to the uniform limit] Let $a,b \in (x_0-R,x_0+R)$. Choose a compact interval $K \subset (x_0-R,x_0+R)$ containing $a$ and $b$. With $S_N$ as above, uniform convergence $S_N\to f$ on $K$ gives \begin{align*} \int_a^b f(x)\,d\mathcal{L}^1(x) = \lim_{N\to\infty}\int_a^b S_N(x)\,d\mathcal{L}^1(x). \end{align*} For each $N$, \begin{align*} \int_a^b S_N(x)\,d\mathcal{L}^1(x) &= \int_a^b \sum_{n=0}^{N} c_n(x-x_0)^n\,d\mathcal{L}^1(x) \\ &= \sum_{n=0}^{N} c_n\int_a^b (x-x_0)^n\,d\mathcal{L}^1(x) \\ &= \sum_{n=0}^{N} c_n\frac{(b-x_0)^{n+1}-(a-x_0)^{n+1}}{n+1}. \end{align*} Passing $N\to\infty$ gives \begin{align*} \int_a^b f(x)\,d\mathcal{L}^1(x) = \sum_{n=0}^{\infty} c_n\frac{(b-x_0)^{n+1}-(a-x_0)^{n+1}}{n+1}. \end{align*} This is the asserted termwise integration formula. [/step]