TITLE: Swapping an improper integral and series
QUESTION [6 upvotes]: What are the hypothesis that allow me to write the identity
$$\int_{x_0}^{+\infty}\left(\sum_{n=1}^{\infty}f_n(x)\right)\mathbf{d}x=\sum_{n=1}^{+\infty}\left(\int_{x_0}^{+\infty}f_n(x)\mathbf{d}x\right)$$
, where $x_0\in\mathbb{R}$ and $\forall n\in\mathbb{N}_{>0}\ \ f_n:[x_0,+\infty)\to\mathbb{R}$ is a continuous function?
I know that exist many theorems about a similar argument, but in the hypothesis the set of integration is a bounded closed interval. Maybe I can use the definition of improper integral? I mean. Fix $M>x_0$, and suppose that $\sum_{n=1}^{+\infty}f_n(x)$ is uniformly convergent over $[x_0,M]$ for all $M>x_0$, then:
$$\lim_{M\to+\infty}\int_{x_0}^{M}\left(\sum_{n=1}^{\infty}f_n(x)\right)\mathbf{d}x=\lim_{M\to+\infty}\sum_{n=1}^{+\infty}\int_{x_0}^{M}f_n(x)\mathbf{d}x$$
but it's not the same thing...

REPLY [2 votes]: The following claim from [Fich, 518 (p. 697)] concerns, as far as I understood, Riemann integrals. Let the series $\sum_{n=1}^{\infty}f_n(x)$ of positive continuous for $x\ge x_0$ (or for $x_0\le x<x_1$) functions has for this values of $x$ a continuous sum $f(x)$. If the latter is integrable on an interval $[x_0,+\infty]$ (or $[x_0,x_1]$) then on this interval the 
series is termwise intergrable. Instead of the convergence of the sum of the series we can assume the convergense of the series $$\sum_{n=1}^{+\infty}\left(\int_{x_0}^{+\infty}f_n(x)\mathbf{d}x\right) \left[\mbox{ or }  
\sum_{n=1}^{+\infty}\left(\int_{x_0}^{x_1}f_n(x)\mathbf{d}x\right)\right].$$
References
[Fich] Grigorii Fichtenholz Differential and Integral Calculus, vol. II, 7-th edition, M.: Nauka, 1970 (in Russian).