Putting the limit inside an integral of sin(nx) as n goes to infinity A couple of days ago I was convinced that $$\lim_{n \to \infty} \int_a^b \sin(nx) \;dx =  \int_a^b \lim_{n \to \infty}\sin(nx) \;dx,$$
but after reading about uniform and pointwise convergence I became really confused. Apperently you can only put the limit inside the integral if the function converges uniformly, but $\sin(nx)$ surely does not converge at all. 
If I evaluate the first integrals I get:
$$\lim_{n \to \infty} \int_a^b \sin(nx) \;dx = \lim_{n \to \infty} \frac{-cos(nx)}{n} \Biggr|_a^b = 0.$$
I can use the same method to integrate the second integral and it would also be equal to $0$, right? So can I put the limit inside the integral in this specific case? Or is it just coincidence that both integrals are equal to each other?

What is $\lim_n\sin(nx)$? The limit is not pointwise convergent, e.g. $\lim_n \sin(n\pi + \pi/2) = \lim_n (-1)^{n+1}$, which clearly doesn't converge. So how are we supposed to interpret the RHS of your equality? The correct answer is that the equality doesn't hold. Why are you convinced it does?
