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From what I'm guessing a keyhole contour is one that looks like thisenter image description here

and because it can be shown that the contribution from $C_R$ and $C_\epsilon$ vanishes as $R\to \infty$, a Hankel contour looks like this.enter image description here

If so, when I'm normally using a keyhole contour to solve improper integrals with bounds at infinity can I start it with something like this? $$\int_{\mathscr{H}}\frac{\ln z}{z^a(1+z)}\ dz=\int_{0}^{+\infty}\frac{\ln x}{x^a(1+x)}\ dx-\int_{0}^{+\infty}\frac{\ln x +2\pi i}{x^ae^{2\pi ia}(1+x)}\ dx$$ Where $\mathscr{H}$ is a Hankel contour surrounding $[0,\infty)$ in the clockwise sense.

bob
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  • No, not "because" it can be shown; if it can be shown (that the contributions vanish). It's not always true! – FShrike Feb 20 '24 at 23:59

1 Answers1

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Your observation is correct in principle. However, there's a good reason to start from the keyhole contour rather than the Hankel contour - the Hankel contour is not closed. If you wanted to actually evaluate any of the integrals you've written down, you'd have to evaluate the contour integral itself, and for this you'd want to apply tools like the residue theorem. You can do this with a keyhole contour but not directly with a Hankel contour.

Gyu Eun Lee
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  • That makes sense. Do you have any examples (not limited to just evaluating definite integrals) of when you would use a Hankel contour? Thanks in advance. – bob Mar 22 '23 at 04:18
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    The Hankel contour was first introduced, I believe, to study properties of the Gamma function. Wikipedia has an article specifically for the Hankel contour, so you could start there. – Gyu Eun Lee Mar 22 '23 at 15:13
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    @bob: the Hankel contour is used in the analysis of Bessel functions. Here is a posting that explicitly considers it. The Keyhole contour is used in the analysis of the Mellin transform (here is an instance of a computation of the Mellin transform of a rational function at a particular point $a=2/3$) – Mittens Jun 16 '23 at 17:16