Integrate eiz/z 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 −iπ, soTaking imaginary and real parts yieldsThe 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.
PV∫−∞∞xeixdx=iπ.
∫−∞∞xsinxdx=π,PV∫−∞∞xcosxdx=0.
Put z=eiθ, so sinθ=(z−z−1)/(2i) and dθ=dz/(iz). The integral becomesThe poles are −2i/3 and −3i/2; only the first is inside the unit circle. The derivative of the denominator there is 12(−2i/3)+13i=5i. The residue theorem therefore givesThere are no poles on the contour because the real denominator is at least one.
∮∣z∣=16z2+13iz−6dz.
∫02π13+12sinθdθ=52π.
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