Residue theorem

If $f$ is holomorphic on and inside a positively oriented closed contour except for finitely many isolated singularities $a_j$ inside it, then

For $0<a<1$, integrate $e^{az}/(1+e^z)$ around a rectangle of imaginary height $2\pi$. Its single pole at $\pi i$ has residue $-e^{\pi ia}$, the top-edge integral is $-e^{2\pi ia}$ times the bottom-edge integral, and both vertical sides vanish at large real width. The residue theorem gives ${\int }_{\mathbb{R}}{e}^{ax}/(1+e^x)dx=\pi /\sin (\pi a)$. Substitution $t=e^x$ gives the equivalent integral ${\int }_0^{\infty }{t}^{a-1}/(1+t)dt$.

For $a>b>0$, substitution $z=e^{i\theta }$ gives poles at zero and the two roots of $b{z}^2+2az+b$. The inside-pole residues sum to $(a-\sqrt{a^2-b^2})/(i{b}^2)$, soThe double pole at zero is essential; omitting it gives the wrong sign and value.

Set $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$ and $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. In an indented upper semicircle use the principal complex logarithm and $z^{1/2}=e^{\log z/2}$. The negative real boundary contributes $iJ-\pi I$, the positive boundary contributes $J$, and both circular arcs vanish. The pole at $i$ has residue $\pi e^{i\pi /4}/4$. Thus $(1+i)J-\pi I=({\pi }^2/(2\sqrt{2}))(-1+i)$; comparing real and imaginary parts yields the displayed integrals.

Suppose a contour follows the real axis, closes in the upper half-plane, and avoids each simple real pole $a_j$ by a small semicircle above it. Each indentation is clockwise and tends toIf the large arc vanishes and there are no enclosed poles away from the real axis, the Cauchy principal value is therefore

Residues at symmetry-related poles can occur with opposite signs and cancel in the contour sum.

Pairing residues at opposite poles can produce an alternating series that converges in a larger half-plane than the corresponding absolutely summed residue series.