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The recent post here has led to the following question (consider $\cos(sx) = \sum\nolimits_{k = 1}^\infty {\frac{{( - 1)^k (sx)^{2k} }}{{(2k)!}}}$). I find it instructive, whether it will turn out to be easy/well-known or not.

So, suppose that $(f_k)_{k \geq 1}$ is a sequence of continuous functions on a non-compact interval $I$. Suppose that $ \sum\nolimits_{k = 1}^\infty {f_k (x)}$ converges uniformly on every compact subinterval of $I$. Further, suppose that both $\sum\nolimits_{k = 1}^\infty {\int_I {f_k (x)\,{\rm d}x} }$ and $\int_I {[\sum\nolimits_{k = 1}^\infty {f_k (x)]} \,{\rm d}x}$ converge (i.e., are finite).

Under these assumptions, can you find an example where $$ \int_I {\bigg[\sum\limits_{k = 1}^\infty {f_k (x)\bigg]} {\rm d}x} \neq \sum\limits_{k = 1}^\infty {\int_I {f_k (x)\,{\rm d}x} }? $$ Consider also the case where we are only given that the right-hand side converges.

If relevant: What about the case where $\sum\nolimits_{k = 1}^\infty {f_k (x)}$ and its partial sums are analytic functions?

Any relevant discussion is welcome here.

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    Wouldn't this be covered by [Fubini's theorem](http://en.wikipedia.org/wiki/Fubini%27s_theorem)?2010-12-19
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    How, for example, would you apply it for $f_k {(x)} = (-1)^k x^k e^{-x} / (k+1)!$, with $I=[0,\infty)$.2010-12-19
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    Are the various convergences assumed to be absolute?2010-12-19
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    @Willie: No....2010-12-19

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Here I use notation $\chi_N$ to be the characteristic function of $[N,N+1]$.

Let $f_1 = \chi_1$. Let $f_k = \chi_k - \chi_{k-1}$ for $k\geq 2$.

On any compact set of the reals, $\sum f_i$ converges uniformly to $f = 0$. So $\int \sum f_i = 0$. Yet $\int f_1 = 1$ and $\int f_k = 0$ for $k \geq 2$. So $\sum \int f_i = 1\neq \int\sum f_i$.


Analyticity doesn't matter. Let $g(x)$ be any arbitrary, positive, integrable function on the real line which decays faster then, say, $|x|^{-1}$. Then if we take $f_1(x) = g(x-1) |_{\mathbb{R}_+}$ and $f_k(x) = g(x-k) - g(x-k+1) |_{\mathbb{R}_+}$ for $k\geq 2$. You get the same result that on any compact subset of the reals $\sum f_i$ converges uniformly to 0. Yet $\sum\int f_i \geq \int_0^\infty g(x) dx$. You can no insert in an arbitrary analytic $g(x)$ (say the Gaussian for example) and get the same conclusion.

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    @Jonas: ah, I missed the continuous assumption. I added a second half to show how this argument can be generalized to be based on (roughly speaking) any integrable function $g$, instead of the characteristic function of an interval. So I can even allow real analytic functions :)2010-12-19
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    Yes, shortly after I posted my comment you added the second part, which made it even more moot than it already was. So I deleted it.2010-12-19