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A contour deformation changes the path of a contour integral through a region where its integrand is a holomorphic function. The Cauchy integral theorem preserves the integral when the endpoints are fixed and the connecting boundary terms vanish. Crossing a pole instead contributes the appropriately oriented residue; a branch cut limits which deformations are permitted on one analytic sheet.

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.

Abel summation applied to the counting function $\lfloor w\rfloor$ gives the displayed identity for $\Re s>1$ and every $x>0$. Since $0\le \{w\}<1$, the integral is locally uniformly convergent and holomorphic for $\Re s>0$. This supplies a meromorphic continuation of the Riemann zeta function to that half-plane, with its sole pole at one and residue one. Keeping the fractional endpoint term makes the formula valid at noninteger cutoffs.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

Let $g$ be a meromorphic function with simple poles ${\alpha }_{+}$ above $c$ and ${\alpha }_{-}$ below $-c$, and $zg(z)\to 0$ uniformly at infinity in $\mathrm{Im}z\ge -c$. The left-to-right line integrals $H(\lambda )=(2\pi i{)}^{-1}{\int }_{\mathbb{R}-ic}g(z)/(z-\lambda )dz$ and $J(\lambda )=-(2\pi i{)}^{-1}{\int }_{\mathbb{R}+ic}g(z)/(z-\lambda )dz$ satisfy $H=g-r_{+}/(\lambda -{\alpha }_{+})$ above $-c$ and $J=r_{+}/(\lambda -{\alpha }_{+})$ below $c$, where $r_{+}$ is the upper residue. Upper semicircle closure proves both formulas by the residue theorem; subtracting the principal part makes $H$ holomorphic at ${\alpha }_{+}$ too. Consequently $H+J=g$ on the intervening strip. No lower-half-plane decay is needed.

The pole at $s=0$ has residue $-(t+2)=-(t-u)/2$ for unit-slope tachyons. It is proportional to conserved scalar-pair vector currents and linear in the scattering-angle cosine, exhibiting a massless spin-one intermediate string state.

For an even field between rigid plates at $y=\pm b$, the transverse modes are $\cos (n\pi y/b)$, with longitudinal wave numbers $k_n=\sqrt{k_0^2-(n\pi /b{)}^2}$. With time dependence $e^{i\omega t}$, outgoing modes into $x>0$ use $e^{-i{k}_{n}x}$: choose positive real $k_n$ for cut-on modes and negative imaginary $k_n$ for cut-off modes. A lower-half-plane Fourier inversion extracts these modes by residues. Their decaying components are evanescent waves, not backward-propagating waves. At an exact cutoff one takes the radiation continuation before its limit.