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The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.

Write $\lambda =-m$ with $m\in \mathbb{N}$. The amplitude is now the rational functionChoose ${\gamma }_1$ to be a small positively oriented circle around the pole $t=1$, and choose ${\gamma }_2$ from $t=0$ along the negative real axis to $-\infty$. At zero the endpoint factor behaves as $t^m$ and vanishes; at negative infinity the exponential decays.

By the residue theorem, the finite-contour solution isEvery derivative term contains $e^z$ times a power of $z$ of degree at most $m-1$, and the highest-degree term is nonzero. Henceup to a constant, where $P_{m-1}$ is a polynomial of degree $m-1$. This is the negative-integer case of the Integer-parameter Laguerre contour residues.

For bosonic Matsubara frequencies ${\omega }_n=2\pi nT$, set $a=vk/(2\pi T)$. Applying the residue theorem to $\pi \coth (\pi z)/(z^2+a^2)$, whose integer poles reproduce the desired summands, givesIt follows thatUsing the area $S_{d-1}=2{\pi }^{d/2}/\Gamma (d/2)$ of the unit sphere, the thermal phase fluctuation becomesAt small $k$, $\coth (vk/(2T))\sim 2T/(vk)$, so the infrared divergence is governed by ${\int }_{0}dk{k}^{d-3}$. It diverges for $d=1,2$ and is finite for $d=3$. Therefore short-range systems cannot have true finite-temperature breaking of this continuous symmetry in one or two dimensions, in agreement with the Mermin-Wagner theorem, whereas it is allowed in three dimensions. In two dimensions a Berezinskii–Kosterlitz–Thouless transition may still produce quasi-long-range order.

If $f$ had a pole at one of the finitely many candidate points, choose a polynomial vanishing to sufficiently high order at all the other candidates. Hermite interpolation lets its remaining Taylor coefficients be chosen so that the residue of $p^{2}f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero integral by the residue theorem, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

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 $P(z)=a_N{z}^N+\cdots +a_0$ with $N\ge 1$ and $a_N\ne 0$, the unique retarded fundamental solution is $E=Hu$, where $P(D)u=0$, $u^{(j)}(0)=0$ for $j<N-1$, and $u^{(N-1)}(0)=1/a_N$. The distributional jump formula for a Heaviside product gives $P(D)E={\delta }_0$. Its explicit expression isThe residue theorem gives the initial derivatives from the coefficients at infinity. Equivalently, the inverse Fourier transform is integrated below all its poles, producing support of a distribution in $[0,\infty )$. Uniqueness follows because the difference of two retarded solutions is an analytic homogeneous solution vanishing on the negative half-line.

With ${\Delta }_F=⟨0|T\{\varphi (x)\varphi (y)\}|0⟩$, for a mode of positive energy $E>0$ the Feynman i-epsilon prescription puts poles $+E-i0$ and $-E+i0$ in opposite half-planes. The energy contour integral closes clockwise below for positive time separation and counterclockwise above for negative separation. The residue theorem gives $e^{-iE|t|}/(2E)$ in either case. Thus the four-dimensional Fourier integral has numerator $i$ and denominator $p^2-m^2+i0$. The integral is understood as a distribution limit, not an ordinary absolutely convergent integral; changing the definition to $i{\Delta }_F$ would change the displayed normalization.

For a nonzero elliptic function with period lattice $\Lambda$, its zeros of a function $a_j$ and poles $b_j$, repeated with their multiplicities, have equal total number and satisfy ${\sum }_j{a}_j-{\sum }_j{b}_j\in \Lambda$. Choose a fundamental parallelogram avoiding all zeros and poles on its boundary. The argument principle applied to $f^{\prime }/f$ gives equality of the numbers. Applying the residue theorem to $z{f}^{\prime }/f$ gives their difference of sums. Pairing opposite edges, the extra factors are the two lattice generators multiplied by integrals of $f^{\prime }/f$ along an edge. These integrals are integer multiples of $2\pi i$, because the endpoints have the same nonzero function value. Thus the difference of sums is a lattice element. This links zeros of a pulled-back line to the chord-and-tangent group law.