Uniform convergence and integrals. I'm asked to tell if the following integral is finite:
$$\int_0^1 \left(\sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right)x^n \right)dx$$
I studied the series (which converges uniformly on $(-1,1)$ by d'Alembert's Criterion and in $-1$ by Leibniz's Criterion, so in general the convergence is uniform in $[-1,1)$).
In $1$ we have that the series goes like $\frac{1}{n}$ and so diverges.
Can I exchange integral and sum if the convergence is not uniform in $1$? I'd say yes because I can write $\int_0^1$ as $\lim_{\epsilon \to 1} \int_0^{\epsilon}$ but I'd like a confirmation.

We know that $f_{n}(x)= \sin \Big( \frac{1}{n} \Big) x^{n}$ are Lebesgue integrable on $(0,1)$ and positive. Since $$ \sum_{n=1}^{ \infty} \int_{0}^{1} \sin \Big( \frac{1}{n} \Big) x^{n} dx = \sum_{n=1}^{ \infty} \sin \Big( \frac{1}{n} \Big) \frac{1}{n+1}, $$ which converges, because $$ \sin \Big( \frac{1}{n} \Big) \leq \frac{1}{n} \Rightarrow \sum_{n=1}^{ \infty} \sin \Big( \frac{1}{n} \Big) \frac{1}{n+1} \leq \sum_{n=1}^{ \infty} \frac{1}{n} \frac{1}{n+1} < \infty $$ we then know by Levi's Theorem for Series of Lebesgue Integrable Functions that $$ \int_{0}^{1} \sum_{n=1}^{ \infty} \sin \Big( \frac{1}{n} \Big) x^{n}  dx =  \sum_{n=1}^{ \infty} \int_{0}^{1} \sin \Big( \frac{1}{n} \Big) x^{n}  dx$$
