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  • $\begingroup$ Just to compare with Integrate[ Exp[-I*(\[Omega] - \[Omega]0)* t]/(\[Lambda]^2 + (\[Omega] - \[Omega]0)^2), {\[Omega], -Infinity, Infinity}, Assumptions -> t \[Element] Reals && \[Omega]0 \[Element] Reals && \[Lambda] > 0] which outputs a long analytic expression in terms of SinIntegral ans CosIntegral. $\endgroup$ Commented Jul 27, 2024 at 4:53
  • $\begingroup$ As you know, I have a problem with this kind of questions here. These searches for quick and easy answers never work and lead to further confusions of unexperienced pupils or students. Future visitors to this forum will not be able to learn about the MA techniques from this questions unless there are some good and pedagogical answers. Unfortunately, your answer is messy and not quite pedagogical because it does not explain why certain transformations are done. $\endgroup$ Commented Jul 28, 2024 at 14:52
  • $\begingroup$ @yarchik: Can you kindly elaborate "certain transformations", giving us details? I follow a usual approach of complex analysis, integrating over Disk[{0, 0}, R, {Pi, 2*Pi}]. $\endgroup$ Commented Jul 28, 2024 at 15:05
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    $\begingroup$ I am not questioning what you are doing. I am concerned that the OP might not even know that the result does not depend on the choice of contour if the function is analytic. Because if they knew, they would be able to find the function poles and residues and/or ask more precise question. Thus, if you would like to be pedagogic, it would be good to briefly explain the relation between the integral over real axis and the contour integral (j in your notations). Moreover, the half-circle integral will only converge if $t-\tau>0$, you only assume the difference is real. $\endgroup$ Commented Jul 28, 2024 at 15:46
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    $\begingroup$ In this case the correct result is (E^2 \[Pi])/2 + ( E^(-t \[Lambda]) (-1 + E^( 2 t \[Lambda])) \[Pi])/\[Lambda] /. {t -> -1, \[Lambda] -> 2, \[Omega]0 -> 1} ao we deal with a bug in ContourIntegrate for t < 0. $\endgroup$ Commented Jul 28, 2024 at 19:51