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I am asked to show that $G(s)=\int^{\infty}_1 \sum\limits^{\infty}_{n=1}g(ny)y^s\frac{\mathrm{d}y}{y}$, where $g$ is Schwartz, is holomorphic. In order to do this, I want to use Morera theorem and Fubini theorem as in Striking applications of Morera's theorem, where someone showed that Gamma function is holomorphic by using the similar approach.

But I don't know how I can show that the condition for Fubini Theorem holds (in other words, if we exchange the integrand by its absolute value, the evaluation of integral is finite).

  1. Is there a great way to show that $\int_C \int_0^\infty |z^{s-1} e^{-z}|\;\mathrm{d}z \;\mathrm{d}s$ is finite? I know the integrand vanishes as $z$ approaches infinity and maybe I want to find the upper bound for that integrand, but I'm not sure how to do it.

  2. Can I do the same approach on $G(s)$?

Thomas Prévost
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jk001
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    The key-idea is to prove that the function is holomorphic on every compact (it is sufficient to deduce that it is holomorphic everywhere). For example, for the $\Gamma$ function, try to see why you can apply Fubini by restricting to a subset of the form $\lbrace s \in \mathbb{C}, \varepsilon \leq \mathrm{Re}(s) \leq A \rbrace$, with $0 < \varepsilon < A$. – TheSilverDoe Feb 18 '21 at 10:30
  • Are you suggesting Weierstrass theorem before using Morera and Fubini, so that I can restrict it to compact subset of $s$? Oh sorry, maybe not. I was confused – jk001 Feb 18 '21 at 10:38
  • No, I am just saying that a function is holomorphic in a domain if and only if it is holomorphic on every compact subset of it. There is no Weierstrass here. And the fact that you can restrict yourself to a compact subset is very useful, it allows you to make easy majorations, in order to apply Fubini for example. – TheSilverDoe Feb 18 '21 at 10:42
  • That makes sense! But then since we are integrating over $C$, which is a compact set, aren't we already restricting the domain? I think I still don't quite see it – jk001 Feb 18 '21 at 10:53
  • Are you suggesting that we can restrict the size of $C$? – jk001 Feb 18 '21 at 10:55
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    No, I am talking about the domain of the function, i.e. in which lives the variable $s$ ! For example, if you want to prove that $\Gamma$ is holomorphic over $\lbrace s \in \mathbb{C}, \mathrm{Re}(s) > 0 \rbrace$, prove that it is holomorphic over the subsets $\lbrace s \in \mathbb{C}, \varepsilon \leq \mathrm{Re}(s) \leq A \rbrace$. This has nothing to do with the path or the interval over which you are integrating. – TheSilverDoe Feb 18 '21 at 10:59
  • Thank you, I think that makes sense. The problem was that I was woefully trying to use the theorem without properly understanding it. – jk001 Feb 18 '21 at 11:15
  • Sorry to bother you, but do you know Fubini's theorem for Riemann integral of complex functions? It seems that Fubini's theorem is the only way to go and I was reading this article https://en.wikipedia.org/wiki/Fubini%27s_theorem but the idea of measure spaces is very exotic to me..... – jk001 Feb 18 '21 at 12:20
  • Fubini's theorem appears in a very natural way (and can be proved more easily) with the formalism of Lebesgue's integral. If you don't know yet the theory of Lebesgue, don't worry : Fubini's theorem is still true with Riemann's formalism, so you can use it with Riemann integral. – TheSilverDoe Feb 18 '21 at 12:48
  • Ok, thanks so much...Does this theorem imply that if the integrand is always positive, I can always interchange the integral without worrying about its convergence? Because if either of them converges, they will be the same. Sorry if this is a stupid question but now I'm very confused why I bothered about the condition of Fubini's theorem in the first place because gamma is always positive when the real part is greater than 1.. And wonder how your hint is relevant – jk001 Feb 18 '21 at 13:03
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    Yes, this is Fubini-Tonelli's theorem, all is automatically ok if the integrand is positive. Be careful, the integrand of $\Gamma$ takes complex values, so it has no sense to say that it is positive... – TheSilverDoe Feb 18 '21 at 13:30

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