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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 $a=0$?

FD_bfa
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Gary
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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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    @RounakSarkar What do you expect from me to do with it? – Gary Oct 12 '21 at 07:52
  • 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
  • @Gary Have you come across an answer to this question? – bob Mar 22 '23 at 05:53
  • @bob No, unfortunately I have not. – Gary Mar 22 '23 at 06:41
  • 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
  • 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
  • Can you please give more details? I might have missed your point. – MathArt Apr 26 '24 at 09:05

1 Answers1

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Edit($12/23/2025$) I deleted my previous answer on $12/12/2025$ (which was my second answer after having deleted a fully incorrect first answer on $12/11/2025$) because I wanted to reconcile the differences between the Wolfram expression (entire function) and the text book expression (has a branch branch point at origin). Now I think I am convinced that my second answer is good and want to undelete. The part that convinced me is: For $\mathfrak{R}(z) \gt 0$, assuming that the branch cut of $\sqrt{z^2}$ is on the imaginary axis, $\operatorname{C}(\sqrt{z^2})$ is indeed the same as $\operatorname{C}(z)$, reconciling Wolfram answer and the text book answer.

Previous answer ($12/12/2025$): I deleted my earlier incorrect answer (there was a delete vote on it from someone which changed my decision to keep it). Here is another attempt.

Let

$$f(t) = e^{t} \sin(t^2)$$

$$ \begin{align} F(s) &= \int\limits_{0}^{\infty} f(t) e^{-st} dt \\ &= \int\limits_{0}^{\infty} e^{t} \sin(t^2) e^{-st} dt \\ &= \frac12 \sqrt{\frac{\pi}{2}}\left( \cos\left( \frac{(s-1)^2}{4} \right) \left( 1-2\operatorname{C}\left( \frac{s-1}{\sqrt{2\pi}} \right) \right) + \sin\left( \frac{(s-1)^2}{4} \right) \left( 1-2\operatorname{S}\left( \frac{s-1}{\sqrt{2\pi}} \right) \right)\right) ; \color{red}{\mathfrak{R}(s) \gt 1} \\ \end{align} $$

Here $\operatorname{C}$ is the Fresnel Cosine integral and $\operatorname{S}$ is the Fresnel Sine integral.

References for the result above:

  1. From Mathematica (Wolfram Cloud to be accurate): wolfram output

  2. From Table of Integral transforms Volume $1$ by Harry Bateman (Section $4.7$, Trigonometric Functions, Page 153, Formula $30$ in my copy) Integral Transforms book

Notes:

  1. The second reference is for $f(t) = \sin(t^2)$, which after multiplying by $e^t$ pushes the ROC to $\mathfrak{R}(s) \gt 1$. The same is seen directly from Wolfram/Mathematica result (which after doing a trivial trigonometric simplification of $\sin^2(p)+\cos^2(p) = 1$ results in exactly the same result as in the book).

  2. $\operatorname{FresnelC}(z)$, $\operatorname{FresnelS}(z)$, $\cos(z^2)$ and $\sin(z^2)$ are all $\color{blue}{\text{entire}}$ functions.

  3. Further proof that $F(s)$ is entire can be seen from Wolfram Alpha singularity analysis:

WA1 WA2 WA3

Conclusion

I think this concludes that we can have a $F(s)$ with analycity on $\mathfrak{R}(s) \gt 0$ that represents a Laplace Transform with ROC $\mathfrak{R}(s) \gt a \gt 0$

Srini
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  • I just noticed that the argument of the fresnel terms are different between the book and Wolfram, which needs some investigation as to who's right, but regardless, either of them are entire functions. – Srini Dec 12 '25 at 21:18
  • I investigated the discrepancy and found out that the book defines $C(x) = \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{x} \frac{\cos(t)}{\sqrt{t}} dt$. This is frustrating as the book seems to be teling (with scaling factors ignored) $FresnelC[\sqrt(z^2)]$ which will have branch point. Only the wolfram result will consolidate my conclusion. Maybe wolfram is wrong and the book is right. In that case, I will delete my answer yet again. Anyone who can shed light on this is welcome. – Srini Dec 12 '25 at 21:52
  • I am deleting this for what it's worth. I promise, I am not returning to this question unless I have a foof-proof counter-example. – Srini Dec 12 '25 at 22:07