Let $f(z,s)$ be a complex-valued function in two complex variables, entire in $s$, and let $$F(s)=\int_\Omega f(x+iy,s)\,dx\,dy$$ where $\Omega$ is an open subset of $\mathbb{C}$. Assume that the integral is known to converge for $s=0$ and any $s\in V=\{s:\Re(s)>1\}$, while we cannot say anything about the remaining domain (might be convergent or not).

Is $F(s)$ analytic (at least on $V$)?

I would say yes: around any point $s_0\in V$ we have a uniform convergent power series $f(z,s)=\sum f_n(z)(s-s_0)^n$, so that $$F(s)=\sum \left(\int_\Omega f_n(x+iy)\,dx\,dy\right)(s-s_0)^n$$ where the exchange of sum and integral is justified by the uniform convergence and the series is convergent (because so is the integral defining $F(s)$ in $V$), showing that $F(s)$ is analytic around any $s_0\in V$.

Assume now that $G(s)$ is an entire function which coincides with $F(s)$ on $V$: it is therefore its unique analytic continuation to the whole of $\mathbb{C}$.

Can we say that $G(0)=F(0)$? That is, does the analytic continuation of $F$ have to agree with the original definition of $F$ at $s=0$, where we know the integral to converge?

The analytic continuation theorem itself does not guarantee this, as $V$ and $\{0\}$ are disconnected (the latter not even being open) but I cannot see how $G(0)$ could differ from $F(0)$, since $F(0)$ and $F(s)$ for $s\in V$ are defined by the exact same integral.

  • $\begingroup$ By saying the integral converges, do you mean $$ \int\limits_{(x,y)\,\in\,\Omega} \left|f(x+iy, s)\right| \, d(x,y) < +\infty \text{ ?} $$ $\endgroup$ – Michael Hardy Jun 15 '18 at 14:26
  • $\begingroup$ My first thought is to see if Morera's theorem can be used. $\endgroup$ – Michael Hardy Jun 15 '18 at 14:27
  • $\begingroup$ @MichaelHardy exactly, the integral converges absolutely at $s=0$ and on $V$. $\endgroup$ – Angelo Rendina Jun 15 '18 at 15:31

We already have a power series expansion of $F$ in $V$. The question is whether this power series converges at 0. If we are given such $G$ that coincides with $F$ in $V$, then the power series of $G$ in $V$ is the same power series as the one of $F$. Saying that $G$ is entire means that its power series converges everywhere. Thus, the power series of $F$ converges everywhere and $F(0)$ is given by the same power series (and, in particular, agrees with $G(0)$).

Edit: I realized that maybe I sort of left out the last step. We saw that $$G(s) = \sum_n \int_{\Omega} f_n(x+iy) ~dxdy ~(s-s_0)^n$$ for all $s \in \mathbb{C}$. Knowing that this series converges everywhere and that the integral defining $F$ converges for $s=0$, we can interchange sum and integral back at $s=0$ to obtain $$G(0) = \int_{\Omega} \sum_n f_n(x+iy)(-s_0)^n ~dxdy = \int_{\Omega} f(x+iy,0) ~dxdy = F(0).$$

  • $\begingroup$ Is the uniform convergence in $s$ enough to justify the exchange of sum and integral in $z$? $\endgroup$ – Angelo Rendina Jun 21 '18 at 17:19

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