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I'm on the last question of my homework and it's involving using the residue theory, which I dont really understand, so could somebody lend me a hand?

I have to evaluate the real convergent improper integral below using residue theory:

$$ \int_0^\infty \frac{ \sin \pi x}{x(1-x^2)} \; \textrm{d}x$$

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    There are a few observations that will help here. First, the "singularity" at 0 is removable, and so will not contribute a residue. Second, the function is even, and so it is enough to evaluate the integral from $-\infty$ to $\infty$ and divide by 2. The last observation is that if you take a large semicircle with base $[-n,n]$ on the $x$ axis, the integral along the circular part (whether you take the top or the bottom) will approach zero as $n$ grows (I believe). From here, the problem becomes a standard residue calculus problem.2012-02-27

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