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Inside its radius a real power series may be integrated term by term on every closed subinterval

Statement

Let $f (x) = \sum_{n \geq 0} a_{n} (x - c)^{n}$ have radius $R$ , and define

$G (x) : = \sum_{n = 0}^{\infty} \frac{a_{n}}{ι (n + 1)} (x - c)^{n + 1} .$

For every closed interval $[u, v]$ strictly inside $∣ x - c ∣ < R$ , the function $f$ is Riemann integrable and

$\int_{u}^{v} f (x) d x = G (v) - G (u) = \sum_{n = 0}^{\infty} \frac{a_{n}}{ι (n + 1)} ((v - c)^{n + 1} - (u - c)^{n + 1}) .$

Thus the power series may be integrated term by term, and the antiderivative series has radius $R$ .

Facts & Assumptions

Given: The power-series sum $f$ , its zero-constant-term formal antiderivative $G$ , and a closed interval $[u, v]$ strictly inside the radius.

[L1]

The antiderivative series has radius $R$ (A power series, its formal derivative, and its zero-constant-term formal antiderivative have the same radius of convergence).

[L2]

Both series converge uniformly on $[u, v]$ (A power series converges absolutely and uniformly on every closed interval strictly inside its interval of convergence), and a uniform limit of integrable functions is integrable with the integral equal to the limit of the integrals (A uniform limit of Riemann-integrable functions is Riemann integrable, and its integral is the limit of their integrals).

[L3]

Termwise differentiation applied to $G$ gives $G^{'} = f$ on the open radius interval (Inside its radius a real power series may be differentiated term by term, and the differentiated series has the same radius).

[L4]

If $H (x) = \int_{u}^{x} f$ , then $H^{'} = f$ on $(u, v)$ because $f$ is continuous on $[u, v]$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive).

[L5]

Differentiable functions are continuous, and two continuous functions on an interval with the same derivative differ by a constant (A function differentiable at $c$ is continuous at $c$ , A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

Proof

technique · direct

1.1L1L2L3

By [L1] and [L2], $G$ and $f$ converge uniformly on $[u, v]$ ; in particular $f$ is integrable there, and [L3] gives $G^{'} = f$ .

2.1step 1.1L3L4L5

Define $H (x) : = \int_{u}^{x} f (t) d t$ . The sum $f$ is continuous as a differentiable function by [L3] and [L5], so [L4] gives $H^{'} = f$ on $(u, v)$ , while the integral construction makes $H$ continuous on $[u, v]$ .

3.1step 2.1L5

The functions $H$ and $G$ are continuous and have the same derivative on the interval. By [L5], $H - G$ is constant; evaluating at $u$ gives $H (v) = G (v) - G (u)$ .

4.1step 3.1L2algebra∎

Subtracting the two convergent series for $G (v)$ and $G (u)$ term by term is licensed by their convergence, and gives the displayed series. This is also the limit of the integrals of the polynomial partial sums by [L2].

Inside its radius a real power series may be integrated term by term on every closed subinterval

Statement

Facts & Assumptions

Proof

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