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Good day, please a dude:

I have $w(s)=\sum_{n\geq 1}^{ }e^{-sn^{2}\pi }$, it is analytic for Re(s)> 0.

Why $\int_{1}^{\infty }w(t)(t^{\frac{-1}{2}-s}+t^{s-1})dt=\sum_{n\geq 1}^{ }\int_{1}^{\infty }e^{-tn^{2}\pi }(t^{s-1}+t^{\frac{-1}{2}-s})dt$ is holomorphic?

I think the question as follows, but I'm not sure : $w(t)$ and $t^{s-1}+t^{\frac{-1}{2}-s}$ are holomorphic, then the product is holomorphic. So, I can use the theorem: if f is analytic, then exist g analytic such that g'=f.

Dan Castro
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  • $F(s) = \int_1^\infty w(t) t^{as+b}dt$ is obviously holomorphic since for $t \in [1,\infty)$ : $w(t) t^{as+b}$ is integrable and holomorphic, and its derivative with respect to $s$ too, so that $F(s)$ is continuous and complex differentiable $F'(s) = \int_1^\infty w(t) (a \ln t) t^{as+b}dt$ – reuns Jun 01 '16 at 17:20
  • https://en.wikipedia.org/wiki/Holomorphic_function#Definition "If $f$ is complex differentiable at every point in an open set $U$, we say that $f$ is holomorphic on $U$. " – reuns Jun 01 '16 at 17:26

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Somewhat more generally, suppose $f(t,s)$ is measurable on $[0,\infty) \times U$ (where $U$ is an open subset of $\mathbb C$), analytic as a function of $s \in U$ for all $t \in [0,\infty)$, and $\int_0^\infty |f(t,s)|\; dt$ is bounded on every compact subset of $U$. Then $F(s) = \int_0^\infty f(t,s)\; dt$ is analytic in $U$. This can be proven using Fubini's and Morera's theorems.

Community
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Robert Israel
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  • I don't get the $f(t,s)$ measurable, while $F_a(s) = \int_0^a f(t,s)dt$ converges uniformly on $s \in U$ to $F(s) = \int_0^\infty f(t,s)dt$ is the sufficient condition. And what is amazing in practice is that when it converges uniformly (so that $F(s)$ is holomorphic) then ${F_a}'(s) = \int_0^a \frac{\partial}{\partial s}f(t,s)dt$ also converges uniformly to $F'(s) = \int_0^\infty \frac{\partial}{\partial s}f(t,s)dt$, proved here using the Cauchy integral formula – reuns Jun 01 '16 at 17:58
  • If $f(t,s)$ is not measurable as a function of $t$, you can't even define the integral. – Robert Israel Jun 01 '16 at 19:28
  • But it's not a very severe restriction, because any explicit example you can write down will be measurable. Non-measurable functions inhabit Axiom-of-Choice land: there are models of Zermelo-Frankel set theory without Axiom of Choice in which they don't exist. – Robert Israel Jun 01 '16 at 19:37
  • you meant measurable on $t \in [0,\infty)$, right, as me, that was unclear ($[0,\infty) \times U$). and I wasn't criticizing your answer, I wanted to share with you the answer to the question of two days ago – reuns Jun 01 '16 at 19:41