Swapping an improper integral and series 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...
 A: The Lebesgue dominated convergence theorem should do it. Call $s_k=\sum_{n=1}^k f_n$, then $|s_n|$ is dominated by a constant by uniform convergence of the series. (In fact, all you need is $|s_\infty|<g(x)$ for some integrable functions $g$ on the domain, doesn't even need to be constant. Then we have by the Lebesgue dominated convergence theorem
\begin{align*}
\lim_{k\to\infty}\int_{[x_0,\infty)}s_kdx &= \int_{[x_0,\infty)}\lim_{k\to\infty}s_kdx,\\
\lim_{k\to\infty}\int_{[x_0,\infty)}\sum_{n=1}^k f_ndx &= \int_{[x_0,\infty)}\lim_{k\to\infty}\sum_{n=1}^k f_ndx,\\
\lim_{k\to\infty}\sum_{n=1}^k\int_{[x_0,\infty)} f_ndx &= \int_{[x_0,\infty)}\lim_{k\to\infty}\sum_{n=1}^k f_ndx,\\
\sum_{n=1}^\infty\int_{[x_0,\infty)} f_ndx &= \int_{[x_0,\infty)}\sum_{n=1}^\infty f_ndx,\\
\end{align*}
EDIT: I had this in the back of my head and I've realised I made an error. I said: "$|s_n|$ is dominated by a constant by uniform convergence of the series." Which is true, however a constant is not an integrable function on $[x_0,\infty)$. 
First you need $f_n$ all integrable, otherwise you can't get anywhere with the right hand side of your identity. Suppose $ K = \text{argsup}_k \sup_x \sum_{n=1}^k f_n$, then a dominating function should be $s_K$, the 'biggest' partial sum. This is integrable because it is a finite sum of integrable functions.
A: 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).
