1
$\begingroup$

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.

So,

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

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

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

Hence

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.