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The poles of ${\csc }^{3}z$ inside $|z|=4$ are $z=-\pi ,0,\pi$. Near $z=k\pi$, write $z=k\pi +w$. Its Laurent series beginsso the residue is $(-1{)}^k/2$. The sum of the three residues isThe positively oriented residue theorem now gives

Forboth $e^z-1$ and $\log (1+z)$ have a simple zero at zero. Hence $g$ has a simple pole, and its residue is

For $h(z)=\sin z\sin (1/z)$, multiplication of the two Laurent series shows that all powers are even:There are infinitely many negative powers, so zero is an essential singularity. There is no $z^{-1}$ term, and therefore

The integrand has an order-two pole at $0$ and a simple pole at $2$. Write it near zero as $h(z)/z^2$, whereIts residue at zero isAt $z=2$ the residue isThe residue theorem now gives

Consider the meromorphic functionClose the contour with a large semicircle in the upper half-plane, indenting above the three real pole $-1,0,1$. The large-arc integral vanishes by Jordan's lemma, while the upper-half-plane indentation rule givesThe three residue areTheir sum is $\cos 1-1$, so the residue theorem yields

The real part of the integrand is an odd function, so its full-line principal value is zero. Its imaginary part,is an even function. Taking imaginary parts and halving the full-line integral therefore gives

Close the contour in the upper half-plane. The degree assumption and Jordan lemma make the integral over the large semicircle tend to zero, while the absence of real zeros avoids indentations of the contour. The residue theorem therefore givesRepeated roots are handled by the usual higher-order residue formula.

For $\Re s>1$, absolute convergence permits separation into odd and even terms:Using the defining integral of the Gamma function and the substitution $u=nt$ givesTermwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum isConsequently the Dirichlet eta function satisfies

Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thusprovides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use insteadwith an integer $k$ for which $1-k^{1-s}\ne 0$; the bounded partial sums of $a_n$ give convergence for $\Re s>0$. This shows that those apparent singularities are removable and yields the Analytic continuation of the Riemann zeta function to the right half-plane.

At $s=1$,whereasHenceso $s=1$ is a simple pole and its residue is

The residue is the coefficient $a_{-1}$ in the Laurent expansion. At a double pole, $h(z)=(z-z_0{)}^{2}f(z)$ is holomorphic and $h(z)=h(z_0)+h\prime (z_0)(z-z_0)+\cdots$, so $\mathrm{Res}(f,z_0)=h\prime (z_0)$.

Put $t=s-1$. The Laurent series of the logarithmic derivative at the simple pole of $\zeta$ has the formwhere in fact $c_0=\gamma$. HenceThe coefficient of $t^{-1}$ in their product, and therefore the residue, iswith the constant $A=-1-2c_0$; equivalently, $A=-1-2\gamma$.

For $\Re s>1$, the Dirichlet series multiplication rule and $-{\zeta }^{\prime }/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ giveApply an effective Perron formula on the line $\kappa =1+1/\log x$ and truncate atThe bound from part (c) controls the truncation error.

Use the classical zero-free region of the Riemann zeta functiontogether with ${\zeta }^{\prime }/\zeta (s)\ll {\log }^2(|\Im s|+3)$ there. Contour shifting moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose residue is $x\log x+Ax$ by part (b). On the new contour,and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp (-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,

The Gamma function recurrence gives $\Gamma (s)=\Gamma (s+j+1)/(s(s+1)\cdots (s+j))$. Its numerator equals one at $s=-j$, so the residue there is $(-1{)}^j/j!$ for every nonnegative integer $j$.

Near a one-particle pole, a scattering amplitude factors into amplitudes coupling each side to the intermediate state, summed over its physical polarizations. The pole position gives its squared mass, while its residue encodes its spin and couplings.

A simple pole is a pole of order one. Its residue is the nonzero value of the holomorphic extension of $(z-a)f(z)$ at the pole.

For the unit-modulus hyperbolic scattering block with elementary mass $m$, a pair bound at rapidity difference $iu$ has mass $m_2=2m\cos (u/2)$. Bootstrap fusion with another elementary particle gives a further direct pole at $3iu/2$. It is inside the physical rapidity strip with positive imaginary residue only for $0<u<2\pi /3$. In that range the new bound state has rest-frame constituent rapidities $iu,0,-iu$, so its energy is $m(1+2\cos u)=m\sin (3u/2)/\sin (u/2)$. Charges add at the bound-state vertex. At $u=2\pi /3$ the apparent pole cancels; beyond that point it lies outside the physical strip. The formula is conditional on this particular scattering block and its direct-pole interpretation, not a universal three-body binding law.

For real $u$, this scalar S-matrix block satisfies $S_u(\theta )S_u(-\theta )=1$ and Hermitian analyticity of a two-particle S-matrix. It therefore has modulus one for real rapidity differences. For $0<u<\pi$ its pole at $\theta =iu$ has residue $2i\sin u$ and admits a direct-channel bound-state pole interpretation. Its crossed block is $\cosh [(\theta -iu)/2]/\cosh [(\theta +iu)/2]$. The endpoints $u=0$ and $u=\pi$ are removable degeneracies giving constant amplitudes, so a particle cannot be inferred there from the uncancelled denominator alone. This block specifies an amplitude; a complete theory must also obey the remaining crossing symmetry and bootstrap fusion constraints.

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.

In units ${\alpha }^{\prime }=1$, generic fixed $t$ gives simple poles at $s=n-1$, $n\ge 0$, with residue $-{\prod }_{j=2}^{n+1}(t+j)/n!$. The empty product is one. A residue zero removes that pole, and fixed $t$ must be away from its own channel singularities.

For $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$ and $a+b+c=1$, the amplitude has generic simple poles at $a=-n$. The gamma function residue and recurrence giveThe residue polynomial has degree $2n$, consistent with closed-string exchange up to spin $2n$. When $a=-n+{\delta }_a$ and $b=-m+{\delta }_b$, the reciprocal factor $1/\Gamma (a+b)$ removes a putative double pole: the leading ratio is ${\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)}^2(1/{\delta }_a+1/{\delta }_b)$. Thus channel singularities must be counted using the kinematic constraint, not by multiplying independent numerator poles.

For a period lattice $\Lambda \subset \mathbb{C}$, the Weierstrass elliptic function is the normally convergent lattice sumIt is an even elliptic function with a double pole of principal part $(z-\omega {)}^{-2}$ and zero residue at every point $\omega$ of the period lattice.