I'm asked to compute using complex methods the following integral: $$ I(a)= \int_0^1 \mathrm{d}x \frac{\sqrt{1-x^2}}{x^2-a^2},$$ where $a>1.$

What I know is the following: for $|z|<1,$ the function $$\sqrt{1-z^2}=\sum_{k=0}^\infty \binom{\frac12}{k}(-)^kz^{2k}$$ is analytic; regarding the denominator, we have to worry about poles for $z=\pm a$ and at $\infty.$

My QUESTIONs are the following:

  1. I'm not sure about the best choice of integration path: can anyone help?
  2. morever, is it necessary to cut a branch in the square root to compute this?
  3. is it possible to do this without using complex methods at all?

For 1. and 2. : make the cut along the interval $[-1,1]$ and use a closed path going once above and once below the interval; you get twice the integral you want to compute (find the result using residue theorem - you'll need to include also the residue at infinity)

For 3. - sure, use e.g. Euler's substitution to get an integral of a rational function

  • $\begingroup$ thanks, so the suggested path should pass between say 1 and $a$? $\endgroup$ – jj_p Sep 12 '13 at 6:53
  • $\begingroup$ i.e., a dogbone? $\endgroup$ – jj_p Sep 12 '13 at 16:12
  • $\begingroup$ @Nicolo: yes, that's a good name for the path $\endgroup$ – user8268 Sep 12 '13 at 16:13

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