Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Ahlfors probably states and proves Morera's theorem. That can prove analyticity of the Gamma function and the Riemann zeta function. –  Michael Hardy Nov 5 '12 at 0:44
    
Perhaps consult some previous questions and their answers: math.stackexchange.com/questions/177953 or math.stackexchange.com/questions/81949 or maybe several others –  GEdgar Nov 5 '12 at 1:22
add comment

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
1  
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 5 '12 at 0:58
1  
... 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 5 '12 at 0:58
add comment

Hint Morera's theorem should tell you something.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.