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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.

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  • $\begingroup$ “The integral is divergent.” $e^{-imt}$ does not diverge as $t\to\infty$. It oscillates with a constant complex magnitude of 1. $\endgroup$ – G. Smith Aug 6 '19 at 19:09
  • $\begingroup$ It is divergent, the square root does not have a finite integral. $\endgroup$ – Qurious Spirit Aug 6 '19 at 19:35
  • $\begingroup$ 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. $\endgroup$ – G. Smith Aug 6 '19 at 19:39
  • $\begingroup$ The $e^{-iEt}$ doesn't matter for integrability, please review the definition of what it means for a function to be integrable. $\endgroup$ – Qurious Spirit Aug 6 '19 at 19:43
  • $\begingroup$ I think physicists make sense of this integral using analytic continuation. See functions.wolfram.com/Bessel-TypeFunctions/BesselK/07/… 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. $\endgroup$ – G. Smith Aug 7 '19 at 1:37