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.

• Since all terms are nonnegative on $[0,1]$, you can use the Lebesgue Monotone Convergence Theorem. Jun 1 '20 at 20:44

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 $$\int_{0}^{1} \sum_{n=1}^{ \infty} \sin \Big( \frac{1}{n} \Big) x^{n} dx = \sum_{n=1}^{ \infty} \int_{0}^{1} \sin \Big( \frac{1}{n} \Big) x^{n} dx$$