How would one evaluate $$\int_0^\infty {x^\alpha\over1+x+x^2}dx\qquad |\alpha|<1,\ \alpha\neq0$$ using complex analysis? I've tried Jordan's lemma but can't get the limits to work out. Any hints?

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    $\begingroup$ $z=0$ is a branch point. You need to consider the keyhole contour. See example IV. $\endgroup$ – Mhenni Benghorbal Mar 25 '13 at 14:00

Using the keyhole contour suggested by Mhenni, you will end up traversing along the real axis up and back. The difference is that you have to take into account the $2 \pi$ phase when going back. Thus, when you show that the integral vanishes along the large arc of radius $R$ as $R \rightarrow \infty$ (why?), and the small arc about the origin of radius $\epsilon$ as $\epsilon \rightarrow 0$ (why?), you end up with

$$(1-e^{i 2 \pi \alpha}) \int_0^{\infty} dx \frac{x^{\alpha}}{1+x+x^2}$$

being equal to $i 2 \pi$ times the sum of the residues of the poles of $1+z+z^2$. I imagine you can evaluate this and thereby determine the value of the integral.

One other thing: be sure to define the arguments of the poles such that they are between $0$ and $2 \pi$. This must be so because you have already defined the branch or section of the Riemann surface when you defined the argument of the points going back along the real line as $2 \pi$. This is a major pitfall and source of frustration for people that I see all too often.

  • $\begingroup$ Does this also work for $\alpha=0$? $\endgroup$ – Fabian Mar 25 '13 at 16:20
  • $\begingroup$ No, but for $\alpha=0$, you do not get such a phase change. For that case, you may still use a keyhole contour, but with the integral $$\oint_c dz \frac{\log{z}}{1+z+z^2}$$ $\endgroup$ – Ron Gordon Mar 25 '13 at 16:24
  • $\begingroup$ Thanks, I've never seen keyhole contours before, but seems reasonably straight forward. The next part of the question is to take the limit $\alpha \to 0$ to find the case $\alpha = 0$ $\endgroup$ – user61496 Mar 27 '13 at 10:10

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