Timeline for A particular contour integral
Current License: CC BY-SA 3.0
9 events
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| Sep 29, 2013 at 13:56 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Sep 29, 2013 at 4:54 | comment | added | Alexandre Eremenko | @Carlo: Yes, for t>0. | |
| Sep 28, 2013 at 13:48 | comment | added | Carlo Beenakker | I'm a bit confused: for $t>0$ the integrand is indeed zero, but why would it be zero for $t<0$? Then the contour has to be closed in the lower half-plane, resulting in this sum over residues $r_j$ (which converges for $t<0$). | |
| Sep 28, 2013 at 2:11 | comment | added | Alexandre Eremenko | There is no "pole at infinity". Infinity is an essential singularity. But the function under the integral decreases sufficiently fast in the upper half-plane to close the contour. | |
| Sep 28, 2013 at 2:10 | comment | added | Alexandre Eremenko | Sorry, 0 is indeed removable. Therefore, the integral is 0. | |
| Sep 28, 2013 at 2:09 | history | edited | Alexandre Eremenko | CC BY-SA 3.0 |
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| Sep 27, 2013 at 19:36 | comment | added | DieLuftDerFreiheit | I agree that the contour should be closed in the upper half plane. I now realize that there is a pole at infinity of order 1/2. How do I account for this pole at infinity when summing residues? | |
| Sep 27, 2013 at 19:35 | comment | added | DieLuftDerFreiheit | I disagree (or I'm wrong and am confused): the origin is not a pole because $$\lim_{x\to0} \hat F(x) = \frac{-i x e^{i x t}}{1-\sqrt{-i x} \coth(\sqrt{-i x})} = 3$$ Furthermore, in the Taylor series expansion of $\hat F(x)$, the root in the denominator at the origin cancels. | |
| Sep 27, 2013 at 17:16 | history | answered | Alexandre Eremenko | CC BY-SA 3.0 |