I am a bit unsure when integration is used together with summation. Here is my question:
Find power series for $\int_0^{1} \frac{\sin x}{x}dx$ in the form $\sum_{k=1}^{\infty} a_kx^k$
Here is what I did (I am not sure if what I do below applies for the definite integral as well - the one in the question):
$$\int \frac{\sin x}{x}dx = \int \frac{\sum_{k=0}^{\infty} (-1)^k x^{2k+1}/(2k+1)! }{x}dx = \int \sum_{k=0}^{\infty} (-1)^k x^{2k}/(2k+1)! dx = \\ \sum_{k=0}^{\infty} \int (-1)^k x^{2k}/(2k+1)! dx = \sum_{k=0}^{\infty}\frac{ (-1)^k x^{2k+1}}{(2k+1)(2k+1)!} $$
Changing the limit to $k=1$. $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1} x^{2k-1}}{(2k+1)(2k-1)!} $$
This is correct according to the key, but I am a bit suspicions about the definite integral in the question. It feels as if I have solved this using an indefinite integral, but it appears to be the same as the definite integral from $0 \to1$.
Is the question wrong, or is it something that I did not consider?