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We would like to compute (or bound) the following complex integral:

$$\int\limits_0^{\infty}\left|\int\limits_{-\infty}^{\infty}\frac{e^{its}}{e^s-\lambda}\,ds\right|\,dt$$

where $\lambda \notin S_{\alpha}$, $\omega >0$, $\alpha \in (\omega, \pi)$, and $S_{\alpha}:= \{z\in \mathbb{C}: |\arg{\lambda}|<\alpha \}.$

We think it should be at least bounded by $1/\left|\lambda\right|C'$ where $C'$ is some constant, the residue method has given us nothing.

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    $\begingroup$ Perhaps the inner improper integral diverges at $s=-\infty$ ? $\endgroup$
    – Gerald Edgar
    Commented May 16, 2018 at 0:23
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    $\begingroup$ @DavidG.Stork The problem was with the inner integral, not with the outer one. Besides, how do you know what the OP really had in mind? I am even confused by the meaning of the word "to bound": from above or from below? $\endgroup$
    – fedja
    Commented May 16, 2018 at 1:50
  • $\begingroup$ @fedja: I don't think I changed that aspect of the meaning in my edits. $\endgroup$
    – David G. Stork
    Commented May 16, 2018 at 2:31

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Mathematica is of the opinion that the inner integraL diverges when $t$ is real, and otherwise claims:

$$ \text{ConditionalExpression}\left[\frac{t \lambda^{i t} B_\lambda(1-i t,0)-t \left(\frac{1}{\lambda}\right)^{-i t} B_{\frac{1}{\lambda}}(i t+1,0)+i}{\lambda t},(\Re(\lambda)=1\lor \Re(\lambda)\leq 0\lor \lambda\notin \mathbb{R})\land \Im(t)<0\land \Im(t)+1>0\right] $$

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