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Keyhole Contour Integral

  • Thread starter ryanwilk
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  • #1
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Homework Statement



Consider I = [tex]\int_0^{\infty} dx \frac{\mathrm{ln}(x)}{x^a(1+x)},[/tex] 0<a<1.

a) Calculate [tex]\oint dz \frac{\mathrm{ln}(z)}{z^a(1+z))}[/tex], along a keyhole contour.

b) Split the contour integral into several parts and calculate these parts separately. Compare to the result of (a) and obtain a value for I.

Homework Equations



[tex]\int_0^{\infty} dx \frac{1}{x^a(1+x)} = \frac{\pi}{sin(\pi a)}[/tex]

The Attempt at a Solution



a) Since the contour avoids the pole at z=0, we only consider the pole at z=-1. The residue of this pole is [tex] \lim_{z\to\ {-1}} \bigg[\frac{\mathrm{ln}(z)}{z^a}\bigg] = \frac{\mathrm{ln}(-1)}{(-1)^a} = \frac{\pi i}{(-1)^a}[/tex]. The integral is then [tex]\frac{2 \pi^2}{(-1)^{a+1}}[/tex].

(not 100% sure about this).

b) I think the contributions from the small and large circles are zero so it should just be that the dx integral is half the contour integral but this would give [tex]I = \frac{\pi^2}{(-1)^{a+1}}[/tex], while wolfram alpha gives [tex]I = \pi^2 \mathrm{cot}(\pi a) \mathrm{cosec}(\pi a)[/tex].

Any help would be appreciated.
Thanks!
 
Last edited:

Answers and Replies

  • #2
97
1
What I was recently taught is:
[tex] -4 \pi i \int\limits_0^{\infty} log(x) R(x) + 4 \pi^2 \int\limits_0^{\infty} R(x) = 2 \pi i \sum res ( log^2 (z) R(z) )[/tex]
R(x) being rational function with no poles for x>0 and such that lim x*R(x)=0 when x->infinity. That should work for your function. And since both integrals (log(x) R(x) and R(x)) are purely real, the formule above splits easily into Real and Imaginary part.
 

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