Timeline for How to calculate contour integrals with Mathematica?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2013 at 14:57 | comment | added | Artes | To make your answer reliable you should draw the branch cuts, so that one can understand it unambiguously. I mean that it is not clear wether the both parts of the unit circle are in the same branch, it is even worse since your plot suggests that these parts belong to different branches, then there is no reason to add integrals over different branches. Another, but rather a small problem is that there should be the symbol I rather than i. | |
| Nov 30, 2013 at 2:54 | comment | added | xslittlegrass | @Artes I mean the function goes like $\frac{1}{\sqrt{z}}$ near the poles, so that the contour integration around the poles are zero. And yes I think we can just parametrized the integration around the circle as long as we make all the branches correct. | |
| Nov 30, 2013 at 2:52 | history | edited | xslittlegrass | CC BY-SA 3.0 | added 15 characters in body |
| Nov 29, 2013 at 23:30 | history | edited | Artes | CC BY-SA 3.0 | added 2 characters in body |
| Nov 27, 2013 at 18:33 | history | edited | xslittlegrass | CC BY-SA 3.0 | added 267 characters in body |
| Nov 27, 2013 at 18:23 | history | answered | xslittlegrass | CC BY-SA 3.0 |