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Suppose I define a function using an integral:

$$f(z)=\int_{\mathbb R} g(z,x)\ dx,$$

where $g$ is some function, $z$ is a complex variable, and $x$ is a real variable. Suppose the integral exists for $z\in U$, where $U$ is some open region. What are sufficient conditions on $g$ so that $f$ is analytic here, and why do they suffice?

I looked in Ahlfors but couldn't find anything relevant.

Potato
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  • Ahlfors probably states and proves Morera's theorem. That can prove analyticity of the Gamma function and the Riemann zeta function. – Michael Hardy Nov 05 '12 at 00:44
  • Perhaps consult some previous questions and their answers: http://math.stackexchange.com/questions/177953/ or http://math.stackexchange.com/questions/81949/ or maybe several others – GEdgar Nov 05 '12 at 01:22

2 Answers2

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It suffices that $g$ is analytic in $z \in U$ for each $x\in {\mathbb R}$ and $\int_{\mathbb R} |g(z,x)|\ dx$ is locally uniformly bounded on compact subsets of $U$. For then if $\Gamma$ is any closed triangle in $U$, Fubini's theorem says $\oint_\Gamma f(z) \ dz = \int_{\mathbb R} \oint_\Gamma g(z,x)\ dz\; dx = 0$, and Morera's theorem says $f$ is analytic in $U$.

EDIT: I guess we'd better also assume that $g(z,x)$ is measurable.

Robert Israel
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    In the long run, a person will better preserve their sanity if such questions are construed as asking when an integral of a function-valued function lies again in the same space as the integrands (and, naturally, with other reasonable compatibilities). Although I anticipate that the kind of answer I am about to give is not as "immediate" as probably desired, I do recommend it long-term: in almost all cases I know, an integral of (parametrized) vectors lies again in the same TVS when the integral can be written as a continuous, compactly-supported integral of vector-valued functions, and ... – paul garrett Nov 05 '12 at 00:58
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    ... invoke existence of Gelfand-Pettis (a.k.a. "weak") integrals of such functions, with values in a quasi-complete (locally convex) TVS. – paul garrett Nov 05 '12 at 00:58
  • When you say Fubini's theorem,you mean just the part switching the integrals. But the part where the double integral equals 0 is given by Cauchy's theorem for closed paths on convex regions, right ? In that case we should add that the $U$ region is convex as a condition. – Conjecture Mar 28 '18 at 09:33
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    @PerelMan It is not necessary that $U$ is convex. The paths $\Gamma$ are triangles contained (with their interiors) in $U$. – Robert Israel Mar 28 '18 at 17:30
  • @RobertIsrael Thank you, indeed! It is then Goursat's Theorem – Conjecture Mar 28 '18 at 17:49
  • @RobertIsrael Could you please confirm that only local integrability over compacts of product spaces is enough to use Fubini's theorem? Because in Fubini's theorem wikipedia page, it's about full integrability: https://en.wikipedia.org/wiki/Fubini%27s_theorem – Conjecture Mar 28 '18 at 17:56
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Hint Morera's theorem should tell you something.

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Mhenni Benghorbal
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