Timeline for Principal value of integral
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2013 at 11:00 | vote | accept | Mark | ||
| May 14, 2013 at 19:37 | vote | accept | Mark | ||
| May 15, 2013 at 11:00 | |||||
| May 14, 2013 at 18:15 | comment | added | Carlo Beenakker | I added several more intermediate steps; is it clear now? | |
| May 14, 2013 at 18:10 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
math corrected
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| May 14, 2013 at 17:36 | comment | added | Mark | Could you explain the last equality in detail? | |
| May 14, 2013 at 17:30 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
explanation
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| May 14, 2013 at 17:19 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
derivation
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| May 14, 2013 at 16:06 | comment | added | Carlo Beenakker | the integral can be done as a contour integral in the complex plane, using the definition of principal value which I gave in a comment to your question above (so as the average of a contour closed in the upper and lower half of the complex plane, once picking up the pole at $x=i\epsilon$ and once at $x=-i\epsilon$; I will try to add this calculation to my answer later today. | |
| May 14, 2013 at 15:09 | comment | added | Mark | @Carlo Beenakker: I am grateful to you for your interest to the question and your answer. I vote it up. However, this formula gives the answer, but not the method. Also the passage to the limit as $\epsilon \downarrow 0$ should be more sound because we deal with the principal value of the specific improper integral with $\cot (x)$. As far as I understand it, that notion uses two limits by itself. | |
| May 14, 2013 at 14:12 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
typo
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| May 14, 2013 at 14:04 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |