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I'm sorry for the banality of the question but I was wondering what happens when, trying to compute an integral with the residue theorem, one (or more) of the singularities lies on the real axis.

For example, if I wanted the F-transform of $$f(x)={1 \over (x^2-9)(1+x^2)(4+x^2)}$$ whose singularities are $\pm3 \pm i, \pm 2i$

Everything is fine until I need to consider the poles on the real axis. Since they're on the boundary of the semicircumference, I suppose their residues would be zero for Cauchy's integral theorem?

Thank you

Silence
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  • Related : https://math.stackexchange.com/q/1012815, https://math.stackexchange.com/q/369276, https://math.stackexchange.com/q/3049997 ... – Jean Marie Jan 12 '20 at 15:27
  • That doesn't quite explain how to behave with singularities that are located on the real axis though... nevertheless it was useful, I didn't know that double pole means introducing a derivative either. So thank you! – Silence Jan 12 '20 at 16:00
  • The singularities on the real axis make a non-zero contribution to the integral. You have to use a a vanishingly small semi-circular contour around the poles on the real axis. See this answer which works the problem with 2 complex poles, and one pole on the real axis: https://math.stackexchange.com/a/3086586/441161 – Andy Walls Jan 14 '20 at 17:53

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