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Jun 11, 2023 at 23:47 comment added Michael Engelhardt Certainly, the notation $\int_{-i\infty }^{i\infty } $ on its own is ambiguous, and one can define a left integral version of it, or a right integral version, or a principal value version (which I think your evaluation is related to) ... that's why this is usually accompanied by an explanation of which contour is meant. My understanding is that the standard definition is chosen the way it is in order to properly construct the inverse Mellin transform.
Jun 11, 2023 at 23:15 comment added Dqrksun Ah I see, it is written in the condition, my bad I overlooked it. Still I find it kind of weird to define it this way because if we consider the contour $\gamma_\epsilon$ traversed right hand side instead. The way we write the integral is still $\int_{-i\infty}^{i\infty}$. Yet in this way the integral would yield a different answer
Jun 11, 2023 at 14:01 comment added Michael Engelhardt The contour in the Mellin-Barnes integral is defined to include $\gamma_{\epsilon} $, so you shouldn't subtract it. You just need to evaluate $C-\Gamma $.
Jun 11, 2023 at 4:34 history edited Dqrksun CC BY-SA 4.0
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S Jun 11, 2023 at 4:33 review First questions
Jun 11, 2023 at 6:56
S Jun 11, 2023 at 4:33 history asked Dqrksun CC BY-SA 4.0