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Integration of Power Series 📂Analysis

Integration of Power Series

Summary

Suppose the power series ∑n=0∞cnxn\sum\limits_{n = 0}^{\infty} c_{n}x^{n} converges at ∣x∣<R\left| x \right| \lt R. We then define the function ff as follows:

f(x)=∑n=0∞cnxn∣x∣<R(1) f(x) = \sum\limits_{n = 0}^{\infty} c_{n}x^{n} \qquad \left| x \right| \lt R \tag{1}

Then the function ff is integrable at (−R,R)(-R, R), and its indefinite integral is as follows:

∫f(x)dx=C+∑n=0∞cnn+1xn+1∣x∣<R(2) \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 ff and ∫f\displaystyle \int f is the same.

Explanation

(2)(2) yields a result as if one integrates the infinite terms of (1)(1) term by term. In other words, one can treat the differentiation of the power series as if polynomial functions are being integrated.

∫[∑n=0∞cnxn]dx=∑n=0∞∫cnxndx \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 ∫f\displaystyle \int f is the same as that of ff. 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 {fn:fn is integrable on [a,b]}\left\{ f_{n} : f_{n} \text{ is integrable on } [a, b] \right\} uniformly converge to ff on the interval [a,b][a, b]. Then ff is also integrable on [a,b][a, b] and the following holds.

∫abf(x)dx=∫ablim⁡n→∞fn(x)dx=lim⁡n→∞∫abfn(x)dx \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 fN(x)=∑n=0Ncnxnf_{N}(x) = \sum\limits_{n = 0}^{N} c_{n}x^{n}. Then fNf_{N} uniformly converges to ff on [a,b]⊂(−R,R)[a, b] \subset (-R, R). By the above lemma, the following holds. At ∣x∣<R \left| x \right| \lt R,

∫f(x)dx=∫lim⁡n→∞∑n=0Ncnxndx=lim⁡N→∞∫∑n=0Ncnxndx=C+lim⁡n→∞∑n=0Ncnn+1xn+1=C+∑n=0∞cnn+1xn+1 \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⁡n→∞1nn=1\lim\limits_{n \to \infty}\sqrt[n]{\dfrac{1}{n}} = 1,

lim sup⁡n→∞∣cn∣nn=lim sup⁡n→∞∣cn∣n \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 ∑n=0∞cnn+1(x−a)n+1\sum\limits_{n = 0}^{\infty} \dfrac{c_{n}}{n + 1} (x-a)^{n+1} is the same as that of ff.

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