Compute the integral of x^a / (1+x^2) for x going from 0 to + infinity

  • #1
disregardthat
Science Advisor
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Compute [tex]\int^{\infty}_0 \frac{x^{\alpha}}{1+x^2} dx[/tex] for some [tex]-1<\alpha<1[/tex].

EDIT: This was slightly wrong.

The hint given is that we can integrate from -p to p except for a small semi-circle around 0, and a large semicircle from p to -p, and choose a branch of [tex]z^{\alpha}[/tex]. Wouldn't this in any case be the method for computing [tex]\int^{\infty}_{-\infty} \frac{x^{\alpha}}{1+x^2} dx[/tex]?

I'd appreciate some help in choosing a proper contour. I'd know how to integrate from -infinity to infinity, but from 0 I have no idea. Also, can I choose my branch for the logarithm arbitrarily?
 
Last edited:

Answers and Replies

  • #2
1,101
3


[tex]
\int^{\infty}_{-\infty} \frac{z^{\alpha}}{1+z^2} \, dz =
\int^{\infty}_{0} (z^{\alpha} + (-z)^{\alpha})\frac{1}{1+z^2} \, dz
[/tex]

But
[tex](-z)^{\alpha} = e^{\pi i \alpha}z^{\alpha}[/tex]
so
[tex]\int^{\infty}_{-\infty} \frac{z^{\alpha}}{1+z^2} dz = \left(1 + e^{\pi i \alpha}\right)\int^{\infty}_0} \frac{z^{\alpha}}{1+z^2} dz[/tex]
 

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