12.3 Integration of Uniformly Convergent Series

Uniform convergence is a powerful tool for integrating infinite series. It allows us to swap the order of integration and summation, making complex calculations easier. This concept builds on earlier ideas about convergence, extending them to functions and integrals.

Understanding when and how to apply uniform convergence to integration is crucial. It helps us solve problems involving infinite series and improves our grasp of advanced calculus concepts. Mastering this topic opens doors to more advanced mathematical analysis techniques.

Integration of Uniformly Convergent Series

Theorem Statement and Proof

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• The theorem states that if a series of functions $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly to $f(x)$ on $[a,b]$, then the series can be integrated term by term on $[a,b]$
• Mathematically, if $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly to $f(x)$ on $[a,b]$, then $\int_a^b f(x) dx = \sum_{n=1}^{\infty} \int_a^b f_n(x) dx$
• To prove the theorem, consider the partial sums $S_n(x) = \sum_{k=1}^n f_k(x)$ and their integrals $\int_a^b S_n(x) dx = \sum_{k=1}^n \int_a^b f_k(x) dx$
  • Since $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly to $f(x)$, for any $\varepsilon > 0$, there exists an $N$ such that $|S_n(x) - f(x)| < \varepsilon$ for all $n \geq N$ and all $x \in [a,b]$
  • Integrating the inequality yields $|\int_a^b S_n(x) dx - \int_a^b f(x) dx| \leq \int_a^b |S_n(x) - f(x)| dx < \varepsilon(b-a)$ for all $n \geq N$
  • This proves that $\int_a^b S_n(x) dx$ converges to $\int_a^b f(x) dx$ as $n \rightarrow \infty$, establishing the theorem

Conditions for Term-by-Term Integration

• A uniformly convergent series $\sum_{n=1}^{\infty} f_n(x)$ can be integrated term by term on $[a,b]$ if the following conditions are met:
  • Each function $f_n(x)$ is integrable on $[a,b]$
  • The series $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $[a,b]$
• If these conditions are satisfied, then $\int_a^b f(x) dx = \sum_{n=1}^{\infty} \int_a^b f_n(x) dx$, where $f(x)$ is the limit function of the series
• The uniform convergence of the series is crucial for the interchange of the integral and the sum to be valid
• Examples of series that can be integrated term by term include:
  • Power series within their interval of convergence
  • Fourier series of continuous functions on a closed interval

Term-by-Term Integration of Series

Evaluating Integrals with Uniform Convergence

• To evaluate an integral involving a uniformly convergent series, first check if the series converges uniformly on the given interval
• If the series converges uniformly, integrate the series term by term using the theorem on the integration of uniformly convergent series
• Calculate the integrals of the individual terms $\int_a^b f_n(x) dx$ and find the sum of the resulting series $\sum_{n=1}^{\infty} \int_a^b f_n(x) dx$
• The sum of the integrated terms will equal the integral of the limit function $\int_a^b f(x) dx$
• Example: Evaluate $\int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n^2} dx$
  • The series $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ converges uniformly on $[0,1]$ by the Weierstrass M-test
  • Integrate term by term: $\int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n^2} dx = \sum_{n=1}^{\infty} \int_0^1 \frac{x^n}{n^2} dx = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \frac{\pi^2}{6} - 1$

Uniform Convergence and Interchanging Limits

• Uniform convergence is a sufficient condition for the interchange of limits and integrals
• If a sequence of functions $\{f_n(x)\}$ converges uniformly to $f(x)$ on $[a,b]$, then $\lim_{n \rightarrow \infty} \int_a^b f_n(x) dx = \int_a^b \lim_{n \rightarrow \infty} f_n(x) dx = \int_a^b f(x) dx$
• This property allows for the evaluation of integrals involving limits by first interchanging the limit and the integral and then evaluating the limit
• Without uniform convergence, the interchange of limits and integrals may not be valid, and counterexamples exist where the equality fails to hold
• Example: Consider the sequence of functions $f_n(x) = \frac{nx}{1+n^2x^2}$ on $[0,1]$
  • $\lim_{n \rightarrow \infty} f_n(x) = 0$ for all $x \in [0,1]$, but the convergence is not uniform
  • $\lim_{n \rightarrow \infty} \int_0^1 f_n(x) dx = \frac{\pi}{4}$, while $\int_0^1 \lim_{n \rightarrow \infty} f_n(x) dx = 0$, showing that the interchange of limit and integral is not valid in this case