Keyhole Contour Integral

[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.