I'm currently working with problems regarding uniform convergence. One of the problems I worked with were stated as the following:
Supoose $f_n(x) = \frac{\sin x/n }{x}$ on the interval $I = [1,\infty)$. Show that the sequence of functions converges uniformly to $f(x) = 0$. Then, show that $\lim_{n\rightarrow \infty} \int_{1}^{\infty} f_n(x)dx = \pi/2 \neq \int_{1}^{\infty} f(x)dx$.
I managed to show that the $f_n(x)$ converges uniformly to $f(x) = 0$. However, I always thought that this implies that we can interchange the order of limits for our integration. In other words, we can bring the $\lim_{n\rightarrow \infty}$ inside the integrand, and solve for $\int_{1}^{\infty} f(x)dx$, which in this case would yield us $0$.
In this case however, it seems to be that we get that this integral is equal to $\pi/2$. I'd be glad if you could explain to me why this doesnt work in this case.
Furthermore, I'll also present the solution to my integration. I hope you can give me feedback on that too.
$\lim_{n\rightarrow \infty} \int_{1}^{\infty} f_n(x)dx = \lim_{n\rightarrow \infty} \int_{1}^{\infty} \frac{\sin x/n }{x} dx = \lim_{n\rightarrow \infty} n \int_{1}^{\infty} \frac{\sin x/n }{x/n} dx$
We use the substitution $x/n = u$, and we get:
$\lim_{n\rightarrow \infty} \int_{1/n}^{\infty} \frac{\sin u }{u} du$
Furthermore, it's given that $\int_{0}^{\infty} \frac{\sin x }{x} dx = \pi/2$ and since $1/n$ tends to $0$ for our lower bound in the previous integral, we precisely get that our limit is equal to $\pi/2$.
Thank you.