Suppose that the domain of convergence of the power series $\sum_{k=0}^{\infty} c_{k}x^{k}$ contains the interval $(-r, r)$. Define $$f(x) = \sum_{k=0}^{\infty} c_{k}x^{k} \hspace{1cm} \text{ > if } |x| < r. $$

Let $[a, b] \subseteq (-r, r).$ Prove that

$$\int_{a}^{b} f(x) \mathop{dx} = \sum_{k = 0}^{\infty} \frac{c_{k}}{k + 1}\left(b^{k + 1} - a^{k + 1}\right).$$

Here's the solution I have. It might be wrong because it's not official.

Recall Theorem $5$, which states that if a sequence of integrable functions $\{f_{n} : [a, b] \rightarrow \mathbb{R}\}$ converges uniformly to the function $f : [a, b] \rightarrow \mathbb{R}$, then the limit function is also integrable.


$$\int_{a}^{b} f(x) \mathop{dx} = \lim_{n\to\infty} \int_{a}^{b} \sum_{k = 0}^{n} c_{k}x^{k} = \lim_{n\to\infty} \sum_{k=0}^{n}\int_{a}^{b} c_{k}x^{k} \mathop{dx} = \lim_{n\to\infty} \sum_{k=0}^{n} \left(\frac{c_{k}}{k + 1}\right)\left(b^{k + 1} - a^{k + 1}\right) $$

$$= \sum_{k=0}^{\infty} \left(\frac{c_{k}}{k + 1}\right)\left(b^{k + 1} - a^{k + 1}\right), $$

which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).

My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?

  • $\begingroup$ How can you get the first equality? $\endgroup$ – tonychow0929 Dec 10 '18 at 6:42
  • $\begingroup$ The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right? $\endgroup$ – joseph Dec 10 '18 at 6:42

Notice that $\forall x \in (-r,r),$

$$f(x) = \sum_{k = 0}^{\infty} c_kx^k = \lim_{n \to \infty} \sum_{k = 0}^{n} c_kx^k.$$


$$\int_{a}^{b} f(x) dx = \int_{a}^{b} \lim_{n \to \infty} \sum_{k = 0}^{n}c_kx^k dx.$$

Now apply Theorem $5$ to pull out the limit.

  • $\begingroup$ Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out. $\endgroup$ – joseph Dec 10 '18 at 6:48
  • $\begingroup$ You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs. $\endgroup$ – tonychow0929 Dec 10 '18 at 6:51

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