Could someone remind me why is incorrect to switch an infinite sum and an integral? Could someone jog my memory on this?
The order of operation between an $\int$ and $\sum_{n\in \mathbb{N}}$ is not always interchangable? Note that the sum is an INFINITE sum
Why is it that $\int \sum_{n \in \mathbb{N}} \neq \sum_{n \in \mathbb{N}} \int$
Is the reason because the integral itself is a sum and the order of "summing" actually matters? (I think it's Multivariable calculus related stuff now)
 A: It is easier and more instructive to give a counterexample using sequences: For example $$\lim_{n\to\infty}\int_0^1 n^2x^n(1-x)\,dx=1$$ even though the integrand goes to zero everywhere in [0,1].
To understand what is going on here, note that the function in the integrand has a graph which is a tall, thin peak getting taller and taller and thinner and thinner as $n\to\infty$, while approaching $x=1$ from the left.
Taking differences, you can easily realize the sequence as partial sums of a series, thus providing the counterexample you seek. To be precise, consider $$\lim_{n\to\infty}\int_0^1\sum_{k=0}^{n-1} \bigl((k+1)^2x^{k+1}(1-x)-k^2x^k(1-x)\bigr)\,dx,$$ which is just a difficult way to write the limit above.
A: Correct -- they cannot always be interchanged. For example
$$ \sum_{n=0}^\infty \int_0^{2\pi} \cos(t+n)\,dt = 0$$
but
$$ \int_0^{2\pi} \left(\sum_{n=0}^\infty  \cos(t+n)\right)\,dt $$
doesn't even exist (the sum never converges).
However, if everything converges absolutely, that is, if either of
$$ \sum_n \int |f(n,t)| \,dt \quad\text{or}\quad \int \sum_n |f(n,t)| \,dt $$
exists, then Fubini's theorem guarantees that the summation and the integral can be done in either order.
A: Switching an limit and an infinite sum constitutes the interchange of limit processes.  Said interchanges often yield unexpected results.
