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Let $f:[0,\infty)\to \mathbb{R}$ be a piecewise differentiable function with $f(0)=0$ and $f'(t)$ of bounded variation. Its Laplace transform $\mathcal{L}f$ converges for $\Re s > 0$. Assume it can be continued meromorphically to $\Re s > -\sigma_0$ for some small $\sigma_0>0$, with a pole at $0$ (and nowhere else).

What would be a natural sufficient condition (or even: a necessary and sufficient condition) on $f$ to ensure that $$\mathcal{L}f(\sigma+ i t) \ll_{\sigma} \frac{1}{t^{1+\epsilon}}$$ for $\sigma>-\sigma_0$?

(For $\sigma>0$, such decay (with $\epsilon=1$) follows immediately from the assumption that $f'(t)$ is of bounded variation and $f(0)=0$.)

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  • $\begingroup$ Am I misunderstanding something, or wouldn't a clear answer to this kind of question address Lindelof and such? $\endgroup$ Commented Aug 16, 2019 at 17:54
  • $\begingroup$ Perhaps. Expound at greater length? $\endgroup$ Commented Aug 16, 2019 at 23:57
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    $\begingroup$ Whew, your previous encrypted/keyboard-translated comment was causing me some cognitive dissonance! :) It's a bit late for me here, but/and tomorrow I will say what I meant... surely nothing surprising to you... hence my confusion. Maybe the issue is about the specific exponents. Anyway, I'll give my impression tomorrow. $\endgroup$ Commented Aug 17, 2019 at 0:16

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