How to Calculate a Keyhole Contour Integral for a Fractional Power Function

[ryanwilk]

Homework Statement

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

a) Calculate \oint dz \frac{\mathrm{ln}(z)}{z^a(1+z))}, 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

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

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 \lim_{z\to\ {-1}} \bigg[\frac{\mathrm{ln}(z)}{z^a}\bigg] = \frac{\mathrm{ln}(-1)}{(-1)^a} = \frac{\pi i}{(-1)^a}. The integral is then \frac{2 \pi^2}{(-1)^{a+1}}.

(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 I = \frac{\pi^2}{(-1)^{a+1}}, while wolfram alpha gives I = \pi^2 \mathrm{cot}(\pi a) \mathrm{cosec}(\pi a).

Any help would be appreciated.
Thanks!

 

[irycio]

What I was recently taught is:
-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) )
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.