# Understanding the solution key to a problem which shows that the integral of a sum equals a given value.

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?

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