If an entire holomorphic function $f(z)$ is given by the analytic continuation of $f(x)=\int_\mathbb{R}e^{-ix\xi}\,d\mu(\xi)$ with a finite Borel measure $\mu$ on $\mathbb{R}$, then $g(x):=\int_\mathbb{R_{\geq 0 }}e^{-ix\xi}\,d\mu(\xi)$ extends holomorphically to an entire function?
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$\begingroup$ Under the assumption, for any $y\in\mathbb{R}$, $\int e^{y\xi}\,d\mu(\xi)<\infty$ is ture? $\endgroup$– user509119Commented Aug 11, 2023 at 14:48
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$\begingroup$ Is it true that the analytic continuation (if it exists) must be $f(z)=\int_\mathbb{R}e^{-i z\xi}\,d\mu(\xi), z \in \mathbb C$ ? $\endgroup$– Gerald EdgarCommented Aug 11, 2023 at 17:16
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$\begingroup$ The expression does not make sense if $\int_\mathbb{R}e^{y\xi} e^{-|\xi|}\,d\xi=\infty$. For instance, $\int_\mathbb{R} e^{-i\xi x}e^{-|\xi|}\,d\xi$ extends to $\frac{1}{1+z^2}$. This is not an entire function, but $\int_\mathbb{R}e^{y\xi} e^{-|\xi|}\,d\xi=\infty$ is remarkable. $\endgroup$– user509119Commented Aug 12, 2023 at 0:45
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1 Answer
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Obviously, $g$ can be holomorphically continued to the lower half plane $\mathbb C^-=\{ z:\textrm{Im}\: z<0 \}$. Moreover, this function will be continuous on $\mathbb C^-\cup\mathbb R$. Since we can write $g(x)=f(x)-\int_{(-\infty,0)} e^{-ixt}\, d\mu(t)$, we see that $g$ also has these properties on $\mathbb C^+\cup\mathbb R$. So we now have a continuous function on $\mathbb C$ that is holomorphic on $\mathbb C\setminus\mathbb R$.
Such a function is entire. I think this is a standard exercise in complex analysis courses, but here's an outline of the argument anyway: By Morera's theorem, it suffices to check that $\int_C g(z)\, dz=0$ for every circle $C$. We can approximately decompose $C$ into the parts in the two half planes and complete each of these by a horizontal piece close to the real axis. By continuity, the contributions from these two extra pieces almost cancel each other out.
answered Aug 12, 2023 at 20:18