Suppose that $F: \mathbb{C} \to \mathbb{C}$ is analytic when $\operatorname{Re}(z)>0$. Assume that it is possible to show that $F$ can be represented as a Laplace transform $$ F(z) = \int_0^{ + \infty } {\rm e}^{ - zt} f(t)\,{\rm d}t, $$ for $\operatorname{Re}(z)>a$ with some non-negative constant $a$ and $f$ being analytic in a neighbourhood of the positive real line. Does it follow from the analyticity of $F$ for $\operatorname{Re}(z)>0$ that $a=0$?
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I love how you asked thus question 2 years ago and never touched it ever again. – Rounak Sarkar Oct 12 '21 at 07:35
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2@RounakSarkar What do you expect from me to do with it? – Gary Oct 12 '21 at 07:52
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If I don't get an answer to one of my questions then I actively seek it by giving bounties and cleverly editing the question to bring to more attention. I made the previous comment because I didn't expected such inactivity in a question from a big user like you. – Rounak Sarkar Oct 12 '21 at 08:26
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@Gary Have you come across an answer to this question? – bob Mar 22 '23 at 05:53
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@bob No, unfortunately I have not. – Gary Mar 22 '23 at 06:41
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I note that this is possible when $f(t)$ is non-negative for all $t$ (possibly away from a compact subset of $\mathbb{R}_{>0}$). For Dirichlet series, this is called Landau's theorem, and for reference you may look for Lemma 15.1 of Multiplicative Number Theory I. Classical Theory by Montgomery & Vaughan. I am not sure whether analyticity of $f$ will make it possible or not. – Kangyeon Moon Apr 22 '24 at 06:18
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I dare not to answer questions from such a big user. But would it be due to some convention issue like $\lim_{\varepsilon\rightarrow 0^+}\int_{-\varepsilon}^\infty$? Actually or strictly $$F(z)=\int_{0^-}^{+\infty } {\rm e}^{ - zt} f(t),{\rm d}t.$$ – MathArt Apr 26 '24 at 08:59
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Can you please give more details? I might have missed your point. – MathArt Apr 26 '24 at 09:05