Integrate around an upper semicircle, indenting the origin by a small upper semicircle traversed clockwise. The integrand is holomorphic inside the indented contour. The large arc vanishes by Jordan's lemma, and the small arc contributes , so
Taking imaginary and real parts yields
The first integrand has a removable singularity at zero and its tails converge conditionally by Dirichlet test for improper integrals. The second integral needs a principal value at zero; its zero also follows from oddness with symmetric exclusion. One must not assign an ordinary convergent integral to its singularity.
Put , so and . The integral becomes
The poles are and ; only the first is inside the unit circle. The derivative of the denominator there is . The residue theorem therefore gives
There are no poles on the contour because the real denominator is at least one.

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