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Can anyone help me prove the following equivalent found in Peskin & Schroeder on page 27. $$ \frac{1}{4\pi^2} \int_{m}^{+\infty} dE \sqrt{E^2-m^2} e^{-iEt} \sim_{t \to \infty} e^{-imt}$$

The integral is divergent, I understand that it has to be understood as distribution but still I'm not sure how to prove it. I'm looking for a rigorous proof possibly using distribution theory. I am not convinced by the simple change of variable argument.

Edit: This question is indeed a duplicate, however, the answer provided is not as rigorous as I would have liked (using distribution theory) and the given answer I believe is not correct. , the correct equivalent is $\sim \frac{e^{-imt}}{t \sqrt{t}}$ and not $\sim e^{-imt}$ as claimed in Peskin and Schroeder or $\sim \frac{e^{-imt}}{t}$ as claimed by the answer in the duplicate.

Qurious Spirit
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  • “The integral is divergent.” $e^{-imt}$ does not diverge as $t\to\infty$. It oscillates with a constant complex magnitude of 1. – G. Smith Aug 06 '19 at 19:09
  • It is divergent, the square root does not have a finite integral. – Qurious Spirit Aug 06 '19 at 19:35
  • You are ignoring (a) the $e^{-iEt}$ factor, and (b) the known non-divergent result! So I am not going to argue further with you. – G. Smith Aug 06 '19 at 19:39
  • The $e^{-iEt}$ doesn't matter for integrability, please review the definition of what it means for a function to be integrable. – Qurious Spirit Aug 06 '19 at 19:43
  • I think physicists make sense of this integral using analytic continuation. See http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/ShowAll.html for its relationship to $K_1$. I think the PS result then comes from the asymptotic oscillatory behavior of $K_1$ on the imaginary axis. – G. Smith Aug 07 '19 at 01:37
  • See https://physics.stackexchange.com/questions/230545/asymptotic-behaviour-of-the-propagator-for-a-scalar-field and https://physics.stackexchange.com/questions/307856/bessel-function-representation-of-spacelike-kg-propagator – G. Smith Aug 07 '19 at 01:51

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