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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)=∑n≥0an(x−c)n have radius R, and define

G(x):=∑n=0∞anι(n+1)(x−c)n+1.

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

∫uvf(x) dx=G(v)−G(u)=∑n=0∞anι(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)=∫uxf, 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
1.1

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

L1L2L3
2.1

Define H(x):=∫uxf(t) dt. 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].

step 1.1L3L4L5
3.1

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).

step 2.1L5
4.1

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].

step 3.1L2algebra∎

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