I know that I can use Lebesgue or monotone convergence theorem to exchange limit of partial sums and a Lebesgue integral, given a power series or a generic function series. But in general given a series $\sum_{n=0}^{\infty}a_n$ which converges, and defined $\int_0^\infty\sum_{n=0}^{\infty}a_n f_n(u)du$ with $f_n(u)$ integrable, I was wondering when I could exchange the integration and the series. In particular in the context of Borel summation , given $\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{a_nu^n}{n!}du$, I was wondering how could I demonstrate that if $\sum_{n=0}^{\infty}a_n$ converges, then I can exchange the integral and the series. (I know that for power series $\sum_{n=0}^{\infty}a_n z^n$ the work can be done using the radius of convergence and I can always find a dominant)
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1$\begingroup$ A very general necessary and sufficient condition for the passage to the limit under the integral sign is known: it is not so easy to use, but you can try and see if it can be useful for your problem. $\endgroup$– Daniele TampieriCommented Oct 18, 2020 at 11:24
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2$\begingroup$ First thing I would try is Fubini. This works for $$\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{a_nu^n}{n!}du$$ provided $\sum a_n$ converges absolutely. $\endgroup$– Gerald EdgarCommented Oct 18, 2020 at 11:52
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As suggested by Gerald Edgar, we can use the Fubini--Tonelli theorem. By the Tonelli theorem, $$\int_0^\infty \sum_{n=0}^{\infty}\Big|e^{-u} \frac{a_nu^n}{n!}\Big|\,du =\sum_{n=0}^{\infty}\frac{|a_n|}{n!}\int_0^\infty e^{-u} u^n\,du =\sum_{n=0}^{\infty}|a_n|<\infty.$$ So, the Fubini theorem is applicable, that is, one can interchange the integral and the series.
answered Oct 18, 2020 at 12:06
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1$\begingroup$ Being pedantic technically we should write the series as an integral over an appropriate measure and moreover we should demonstrate that $$\int_0^\infty \sum_{n=0}^{\infty}\Big| e^{-u}\frac{a_nu^n}{n!}\Big|\,du < \infty.$$ and because of integrability, we can swap the oringinal integral and sum. Now here $\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{|a_n|u^n}{n!}\,du =\sum_{n=0}^{\infty}\frac{|a_n|}{n!}\int_0^\infty e^{-u} u^n\,du$ are we using Tonelli's Thm. for non-negative measurable functions? But if the series is uniformly convergent can we use the Dominated Convergence Theorem? $\endgroup$ Commented Oct 18, 2020 at 12:56
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1$\begingroup$ @Coltrane8 : (i) Of course, the sum is the integral over the counting measure. (ii) I have now taken into account your point about the use of the Fubini theorem. (iii) I don't see a way to use dominated convergence. $\endgroup$ Commented Oct 18, 2020 at 13:33