The Definite Integration of a discrete sum is equivalent to the discrete sum of a Definite Integration? I am exploring a famous definite integral that is critical to the Blackbody radiation problem. The solution to the integral is laid out in great detail here:
http://www.sosmath.com/CBB/viewtopic.php?p=67530
I understand every point proffered by the poster except one. How can the order of the discrete sum operator and the Integral operator be exchange without breaking the equivalence relation? Considering that both operators are linear operators, I can intuit that order may not matter. However, the fact that the Sigma operator sums over discrete quantities while the Integral sums continuously over infinitesimal quantities causes me to doubt whether the order of both operators can be exchanged while maintaining equality. Time constraints force me to accept on faith that this is true but I am eager to look into the validity of this. Can anyone point me to some mathematical sources that will elucidate this matter? Thank You. 
 A: (Expanding on RRL's comment)

To really understand what's happening, we need to be explicit about what we mean when we write down an infinite sum of the form $\sum_{n=1}^\infty$.

In particular,
\begin{align*}
\alpha & =\int_{0}^{\infty}\frac{x^{3}}{e^{x}-1}dx\\
 & =\int_{0}^{\infty}x^{3}\lim_{N\rightarrow\infty}\sum_{n=1}^{N}e^{-nx}dx\\
 & =\int_{0}^{\infty}\lim_{N\rightarrow\infty}x^{3}\sum_{n=1}^{N}e^{-nx}dx.
\end{align*}
We would like to, at this point, "pull" the $\lim_{N}$ "out" of the integral. To do this, note that the sequence
$$
\left\{ \textstyle{\sum_{n=1}^{N}}x^{3}e^{-nx}\right\} _{N}
$$
is nonnegative and monotonically increasing, and hence
\begin{align*}
\alpha & =\lim_{N\rightarrow\infty}\int_{0}^{\infty}x^{3}\sum_{n=1}^{N}e^{-nx}dx & \text{(Lebesgue's monotone convergence)}\\
 & =\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\int_{0}^{\infty}x^{3}e^{-nx}dx & \text{(linearity of integral)}\\
 & =6\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{n^{4}}\\
 & =\frac{\pi^{4}}{15}.
\end{align*}
