Continuity and Integrability Any help with this problem is appreciated. 
Consider the function $f(x) = \sum_{n=1}^\infty x n^{-\beta} e^{-nx}$. For what values of $\beta \in \mathbb{R}$ is $f$ continuous on $[0,\infty)$ and in $L^1[0,\infty)$?
 A: I'll use the notation $f_\beta(x)$ for your function, to make explicit the dependence on $\beta$.
First, recall that a series of continuous functions that is uniformly convergent on an interval converges to a continuous function there. You should thereby be able to show that $f_\beta$ is continuous on $(0,\infty)$ for any $\beta\in\mathbb R$. Clearly $f(0)=0$ for any $\beta\in\mathbb R$. So the issue is to decide for which $\beta$ we have $\lim_{x\to0+} f_\beta(x) = 0$, to make it continuous at $0$.
The uniform convergence argument above actually works even at $0$ when $\beta>1$, as you should verify. Also, you can evaluate $f_\beta(x)$ exactly when $\beta=0$ (geometric series), showing that it is not continuous at $x=0$; then a comparison of values should show the same when $\beta<0$. So the crucial interval is $0<\beta\le 1$. I believe it is in fact continuous for these values, but a further argument is needed.
Uniform convergence also allows term-by-term integration of series, so you should be able to relate the integral of $f_\beta$ to functions like $f_{\beta+1}$ and $f_{\beta+2}$, at least on $(0,\infty)$. That should allow you to deduce, from your knowledge from the first part, that $f_\beta$ is integrable exactly for $\beta>-1$. (Again, if you can prove that it's not integrable for $\beta=-1$, then a comparison immediately shows the same for $\beta<-1$.)
