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For $\mathrm{Re}s>1$, define the Riemann zeta function by the absolutely convergent Dirichlet seriesThe Euler product follows from unique prime factorization and absolute convergence. The counting-function integral givesBecause the fractional part is bounded, the last integral, and its derivatives with respect to $s$ on compact subsets, converge uniformly for $\mathrm{Re}s>0$. It is a holomorphic function there. This proves the Meromorphic continuation of the Riemann zeta function to the right half-plane, with a simple pole at one of residue one and no other singularity in that half-plane.

The Gamma function isThe Gamma function recurrence extends it meromorphically; its poles are at the nonpositive integers and it has no zeros. The functional equation of the Riemann zeta function can be stated asor, equivalently,These are identities of meromorphic functions; the second also extends the Riemann zeta function to the remaining half-plane. The critical strip is $0<\mathrm{Re}s<1$; its boundary lines will also be treated below, rather than included among possible exceptional zeros.

For $\mathrm{Re}s>1$, the absolutely convergent reciprocal Euler product gives $\zeta (s)\ne 0$. For $\mathrm{Re}s<0$, the factors $2^s$, ${\pi }^{s-1}$, $\Gamma (1-s)$ and $\zeta (1-s)$ in the second functional equation of the Riemann zeta function are finite and nonzero. Hence the only zeros there are the zeros of the sine factor:Each is simple. At zero, the sine zero cancels the pole of $\zeta (1-s)$; using its residue one gives $\zeta (0)=-1/2$, not zero. Nonvanishing on the rest of $\mathrm{Re}s=0$ follows from the boundary-line proof below and the functional equation of the Riemann zeta function. Together these facts prove that the trivial zeros of the Riemann zeta function are the only zeros outside the open critical strip.

Here is a Jensen disk proof of the zeta zero-count bound which avoids any unproved left-half-plane growth estimate. Set $F(s)=(s-1)\zeta (s)$, a holomorphic function throughout $\mathrm{Re}s>0$, including at one. The integral continuation formula givesFor each integer $j$, apply Jensen's formula in the disk with centre $2+ij$, outer radius $R=7/4$ and inner radius $r=8/5$. The outer disk stays in $\mathrm{Re}s\ge 1/4$, so its maximum modulus is $O((|j|+2{)}^2)$. At its centre,since $|1/\zeta (2+ij)|\le {\sum }_n|\mu (n)|n^{-2}\le \zeta (2)$. Therefore the number of zeros in its inner disk, counted with multiplicity, is at mostThe rectangle $1/2\le \mathrm{Re}s\le 1$, $|\mathrm{Im}s-j|\le 1/2$ fits in the inner disk because its farthest point has distance $\sqrt{(3/2{)}^2+(1/2{)}^2}=\sqrt{5/2}<8/5$. Summing over $O(T)$ such disks counts $O(T\log T)$ zeros in the right half of the critical strip up to height $T$. The first functional equation of the Riemann zeta function bijects zeros, with multiplicities, in the left half with their reflections $s\mapsto 1-s$ in the right half; its Gamma function factors are finite and nonzero in the strip. ThusThe pole of zeta function at one does not count as a zero: $F(1)=1$.

Finally, for real $\sigma >1$ and $t\ne 0$, the Euler-product logarithms and the nonnegative trigonometric polynomialgive the three-four-one product proof of zeta boundary nonvanishing:If $\zeta (1+it)$ had a zero of order $a\ge 1$, its factor would be $O((\sigma -1{)}^{4a})$ as $\sigma \downarrow 1$. The real zeta function factor has pole order three, and the factor at $1+2it$ stays bounded because $t\ne 0$. The product would tend to zero as $O((\sigma -1{)}^{4a-3})$, contradicting that it is at least one. ThereforeAt $t=0$ there is a pole, not a zero. Applying the sine-form functional equation of the Riemann zeta function at $s=it\ne 0$ now proves nonvanishing on the remaining imaginary axis, completing the earlier assertion about all zeros outside the critical strip.

The denominator vanishes when $z=(2j+1)i\pi /2$. For $R>0$, only $z_0=i\pi /2$ lies inside the rectangle. This is a simple pole, since the denominator derivative there is $2$. Its residue isThe positive orientation and the residue theorem consequently give

In elastic diagonal factorized scattering, the Faddeev-Zamolodchikov algebra describes an exchange of two particle operators. Exchanging the pair twice must restore the original state. This is analytic unitarity:Keeping the reversed species in this identity avoids an unstated parity assumption. For the particular amplitudes derived below, crossing and analytic unitarity also imply $S_{\bar{A}A}=S_{A\bar{A}}$. Hermitian analyticity of a two-particle S-matrix states $S_{ab}(\theta {)}^{\ast }=S_{ba}(-{\theta }^{\ast })$ and, together with analytic unitarity, gives $|S_{ab}(\theta )|=1$ for real rapidity differences. Crossing symmetry analytically turns an incoming particle into an outgoing antiparticle, relating the two channels by $S_{A\bar{A}}(\theta )=S_{AA}(i\pi -\theta )$, with the reverse relation obtained by crossing again. The shift $i\pi$ follows from the sign reversal of the two-momentum $p(\theta )=(m\cosh \theta ,m\sinh \theta )$.

Put $u=\lambda /2$. The given amplitude is the unit-modulus hyperbolic scattering block $S_u(\theta )=\sinh [(\theta +iu)/2]/\sinh [(\theta -iu)/2]$. Its numerator and denominator interchange under $\theta \mapsto -\theta$, proving analytic unitarity. For real $u$ it also obeys Hermitian analyticity of a two-particle S-matrix. Applying crossing symmetry givesThis crossed amplitude also has unit modulus on the real axis. Thus both requested channel constraints hold.

The bound-state conclusion needs a coupling range. In the fundamental attractive range $0<u<\pi$, equivalently $0<\lambda <2\pi$, the denominator has a simple pole at $\theta =iu$ inside the physical rapidity strip. Its residue iswith positive imaginary coefficient. Under the usual one-particle interpretation of this direct-channel bound-state pole, it couples two charge-$+1$ particles to a new charge-$+2$ particle $B$. In the center-of-mass frame its constituents have analytically continued rapidities $\pm iu/2$, and their total two-momentum is $(2m\cos (u/2),0)$. Thus the relativistic bound-state mass from a rapidity pole givesIt is positive and less than the two-particle threshold $2m$. Without the coupling qualification the requested deduction is false: at $\lambda =0$ the amplitude is identically one after removing the apparent $0/0$ at the origin. A free massive complex scalar field has this diagonal amplitude, charge-$\pm 1$ particles and no isolated charge-$+2$ bound state. At $u=\pi$ the amplitude similarly becomes constant $-1$ and the apparent boundary pole cancels. Neither endpoint supplies the claimed particle.

For bound-state fusion of factorized S-matrices, represent $B$ as the residue of $A({\theta }_B+iu/2)A({\theta }_B-iu/2)$ at its bound-state separation. Move a third $A({\theta }_A)$ through both constituents using the Faddeev-Zamolodchikov algebra, then take the same residue. The bound-state normalization occurs on both sides and cancels. Consequently bootstrap fusion giveswhere $\theta ={\theta }_B-{\theta }_A$. This has analytic unitarity as a product of two shifted blocks. Its poles occur at $\theta =iu/2$ and $\theta =3iu/2$, modulo $2\pi i$; numerator zeroes occur at the corresponding negative positions, subject to cancellations at special couplings.

The nearer pole, $\theta =iu/2$, has residue $-2i\sin u$. It is a crossed-channel pole in diagonal factorized scattering, not a new direct-channel charge-$+3$ state. Indeed, the exchanged momentum has invariantusing $m_B=2m\cos (u/2)$. The exchanged particle is therefore the already present $A$, with the appropriate charge flow at the crossed vertex. Equivalently, crossing symmetry puts a positive-residue direct pole of $S_{B\bar{A}}$ at $i(\pi -u/2)$, corresponding to $B+\bar{A}⟶A$. This distinction avoids assigning an extra mass by applying the direct-channel formula to every pole.

The farther pole, $\theta =3iu/2$, lies in the physical rapidity strip only for $0<u<2\pi /3$, equivalently $0<\lambda <4\pi /3$. Its residue iswhich has positive imaginary coefficient in that range. It gives a charge-$+3$ bound state $C$. The relativistic bound-state mass from a rapidity pole now yieldsThe positive root in the admitted range isThis is the three-particle bound state from equal-mass fusion; in its own rest frame the constituent rapidities are $iu,0,-iu$. At $u=2\pi /3$ the farther apparent pole cancels because its numerator also vanishes; for $2\pi /3<u<\pi$ it lies outside the physical rapidity strip and does not require a new charge-$+3$ particle. Charge-conjugate partners carry charges $-2$ and $-3$ where the corresponding states exist. Fusion distinguishes an existing crossed-channel particle from a genuinely new direct-channel bound state, and the latter requires the stated smaller coupling range.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

Use the positively oriented unit circle and $z=e^{i\theta }$. Writing $d=\sqrt{a^2-b^2}$, the contour integral for a sine-squared trigonometric quotient becomesThere is a double pole at zero and simple poles at $r_{\pm }=(-a\pm d)/b$. Since $a>b>0$, $|r_{+}|<1$ and $|r_{-}|>1$. The coefficient of $z^{-1}$ at zero is $a/(i{b}^2)$. At $r=r_{+}$, the residue isbecause $(r-r^{-1}{)}^2=4d^2/b^2$. The residue theorem consequently givesFor the second integral, use $x=\cos (\theta /2)$ with $0\le \theta \le \pi$. The square root is $\sin (\theta /2)$, and reversing the integration limits givesThe last factor uses the symmetry of the integrand about $\pi$ over a full period. Therefore

For $\Re s>1$, the Riemann zeta function is the absolutely convergent Dirichlet series $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$. Apply Abel summation to the tail, with counting function $\lfloor w\rfloor -\lfloor x\rfloor$. The upper boundary term tends to zero, givingSubstitute $\lfloor w\rfloor =w-\{w\}$ and integrate the $w$ term. This proves the fractional-part continuation formula for the Riemann zeta functionThe endpoint convention is valid whether or not $x$ is an integer. Since $0\le \{w\}<1$, the last integral converges locally uniformly for $\Re s>0$, including after differentiation on compact subsets. It therefore defines a holomorphic function there. The other terms are entire except for $x^{1-s}/(s-1)$. By the identity theorem for holomorphic functions, the formula supplies a meromorphic continuation with exactly one pole in $\Re s>0$: a simple pole at $s=1$ of residue one.

Let $A(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converge absolutely at a real $c>0$, and take $T\ge \max (2,c)$. DefineThe truncated Perron kernel estimate isFor $v>1$ close the contour to the left, collecting the residue one at zero; for $v<1$ close it to the right, collecting no residue. On the horizontal sides, integrating $v^{\sigma }$ bounds the error by $O(v^c/(T|\log v|))$, and the remote vertical side tends to zero. Near $v=1$ use the bounded transition estimate instead, giving $O(v^c)$. These are the contours and bounds underlying the kernel formula. The constants are uniform for $1\le c\le 2$, the range needed below.

Absolute convergence permits termwise integration. The truncated Perron formula is consequentlyHere the primed sum has half weight when $x$ is an integer, and $a_x=0$ otherwise. To obtain the inclusive sum add $a_x/2$ at an integer. This endpoint convention avoids a false uniform assertion about the kernel at $v=1$.

Apply this with $a_n=\Lambda (n)$, $A=-{\zeta }^{\prime }/\zeta$, $c=1+1/\log x$ and $2\le T\le x$, for sufficiently large $x$. For $n\notin [x/2,2x]$, $|\log (x/n)|$ is bounded below, andusing $-{\zeta }^{\prime }(c)/\zeta (c)\ll (c-1{)}^{-1}$. In the central range $(x/n{)}^c\ll 1$, $\Lambda (n)\ll \log x$, and $|\log (x/n)|\asymp |n-x|/x$. Separate the nearest integers, then sum the harmonic tail over distances $j\ge 1$: its contribution is $O((x/T){\log }^{2}x)$. The possible endpoint weight and nearest terms, of size $O(\log x)$, are absorbed because $T\le x$. ThereforeFor $x+y$, use the same vertical line $c=1+1/\log x$; the error estimate remains valid since $x\le x+y\le 2x$. Subtract the two formulas. The identityhas absolute value $O(y)$ on that line, because $(2x{)}^{c-1}$ is bounded. Taking absolute values givesThis is the short-interval Perron bound for the second Chebyshev function.

Take an oriented closed string in flat, critical bosonic string theory, with ${\alpha }^{\prime }=1/(2\pi T)$ and a mostly-plus target Minkowski metric. After continuation of the worldsheet to Euclidean signature, the Polyakov path integral sums over embeddings $X$ and worldsheet metrics, divided by worldsheet diffeomorphisms and Weyl transformations. In a flat target its kinetic action isTarget-time continuation or analytic continuation of external momenta defines the Lorentzian scattering amplitude; a naive real Euclidean Gaussian for timelike $X^0$ would not be convergent.

An external tachyon is represented by the tachyon vertex operator $V_p(z,\bar{z})=:e^{ip\cdot X}:$, with conformal weights $h=\bar{h}={\alpha }^{\prime }{p}^2/4$. The physical integrated vertex has $(h,\bar{h})=(1,1)$, so $p^2=4/{\alpha }^{\prime }$. Schematically its tree amplitude isThe Faddeev-Popov determinant from conformal gauge is represented by the worldsheet ghost fields. At tree level the worldsheet is a Riemann sphere, whose unpunctured complex structure has no moduli. Its residual conformal automorphisms are the Möbius transformations, $PSL(2,\mathbb{C})$. Fix three insertion points, accompanying their unintegrated vertices by the required $c\tilde{c}$ ghost factors. The remaining $N-3$ complex insertion positions are integrated over the sphere; equivalently one integrates all positions and divides by the residual conformal group.

The embedding fields are free, withTheir zero-mode integral gives momentum conservation, and their nonzero-mode Gaussian integral gives the Koba-Nielsen factorThis explains the sphere tachyon position integral without needing its evaluation. For four tachyons the resulting Virasoro–Shapiro amplitude displays the exchanged string spectrum directly.

Use all-incoming external momenta and introduce $u=-(p_1+p_4{)}^2/(8\pi T)$ alongside the two printed dimensionless Mandelstam variables. Since $p_a^2=8\pi T$ and ${\sum }_a{p}_a=0$, one obtains $s+t+u=-4$. The physical center-of-mass energy squared in the $s$ channel is $s_{\mathrm{phys}}=8\pi Ts=4s/{\alpha }^{\prime }$. The $t$ channel measures the analogous crossed momentum transfer, with sign determined by the mostly-plus convention. Rewriting the Gamma factors in a symmetric form givesThe Gamma function poles imply, at generic fixed values of the other invariant,The denominator Gamma factors can remove residues at special intersecting channel kinematics; the statement concerns a generic single-channel limit. An $s$-channel pole occurs when the intermediate momentum $p_1+p_2$ satisfies the mass-shell condition for a closed-string state:The pole at $s=-1$ exchanges the ground-state tachyon, the pole at $s=0$ exchanges massless states, and the positive integer poles exchange the infinite massive tower. The $t$-channel interpretation is the crossed version. Factorization means that each residue is a sum of products of couplings to intermediate physical states that couple to the chosen external particles. It need not expose every representation at that mass.

For example, the Gamma function recurrence and Gamma function residue at a nonpositive integer give the dimensionless Virasoro–Shapiro amplitude pole residueThe residue polynomial has degree $2n$, consistent with maximum spin $2n$ in the exchanged level. The amplitude also has the corresponding $u$-channel poles by crossing symmetry.

The massless fields can be treated as target backgrounds rather than separate asymptotic insertions. Write the target metric as $G_{mn}={\eta }_{mn}+h_{mn}$, introduce a Kalb–Ramond field $B_{mn}$, and a dilaton $\Phi$. In conventional Euclidean signs their string nonlinear sigma model action isHere ${ϵ}^{ab}$ is the antisymmetric tensor density. The three backgrounds correspond to the graviton, antisymmetric tensor and scalar states at closed-string level one. Expanding the vacuum functional in $h,B,\Phi$, then Fourier expanding the backgrounds, produces exactly their integrated string vertex operators. Its functional derivatives therefore generate the amplitudes with massless external strings. The connected vacuum functional organizes connected amplitudes; the spacetime effective action organizes the corresponding vertices after treating massless propagation consistently.

At momenta small compared with $1/\sqrt{{\alpha }^{\prime }}$, massive string propagators can be expanded in powers of momenta over their masses. The analytic part of the amplitudes consequently determines local higher-derivative interactions, ordered by powers of ${\alpha }^{\prime }$. Massless exchange poles are retained through propagation of the massless fields, rather than expanded into local contact terms. Up to field redefinitions, the leading massless-sector action in the string-frame metric isThe terms denoted $O({\alpha }^{\prime })$ contain additional derivatives, including curvature-squared terms in the bosonic theory. The expansion concerns the massless sector around the perturbative bosonic background; the tachyon instability remains and is not cured by omitting its field from this displayed action. Thus this is a formal perturbative effective description, not a claim of a stable bosonic vacuum.

Finally split the dilaton into a constant and its variation, $\Phi ={\Phi }_0+\varphi$. By the Gauss-Bonnet theorem, its dilaton Euler-characteristic weighting on a connected closed oriented surface follows fromDefine the string coupling by $g_s=e^{{\Phi }_0}$. A genus-$g$ path integral is weighted byThe sphere carries $g_s^{-2}$, the torus carries $g_s^0$, and each extra handle adds $g_s^2$. At higher genus one integrates over complex-structure moduli as well as insertion points, with the associated antighost insertions supplying the correct moduli measure. With canonically normalized external vertices an $N$-point genus-$g$ amplitude scales as $g_s^{2g-2+N}$.

Summing connected worldsheets of every genus of a surface yields the string-loop effective action expansionEach coefficient has its own low-energy ${\alpha }^{\prime }$ expansion. The two parameters have different roles: ${\alpha }^{\prime }$ resolves finite string size through higher derivatives, while $g_s^2$ counts additional string loops.

Suppress the common time factor $e^{i\omega t}$, take $\omega >0$, and fix the spatial Fourier transform conventionA plus transform is supported on $x>0$ and analytic above $\Gamma$; a minus transform is supported on $x<0$ and analytic below it. For the outgoing radiation condition with this time convention, initially take $k_0=\omega /c_0-i\epsilon$, $\epsilon >0$, and pass to the limit at the end. Thus $+k_0$ lies below the contour and $-k_0$ above it.

Choose $\gamma =(k^2-k_0^2{)}^{1/2}$ to have positive real part on the real transform line. Its branch cuts run from $+k_0$ into the lower half-plane and from $-k_0$ into the upper half-plane, without crossing $\Gamma$; downward and upward vertical rays are suitable. In the zero-absorption limit, $\gamma >0$ for real $|k|>k_0$ and $\gamma =+i\sqrt{k_0^2-k^2}$ between the branch points. This ensures that $e^{-\gamma |y|}$ represents decay or outgoing radiation, rather than an incoming exterior field.

Evenness in $y$ and the Helmholtz equation give the transformed fieldsAt $y=b$, the one-sided normal derivatives of the scattered field agree for $x<0$, because that part of the interface is open. For $x>0$ they both vanish by rigidity. Their common trace is therefore a minus function, denoted $V^{-}(k)={\partial }_y{\Phi }^{-}{|}_{y=b}$. ConsequentlyThe jump in the total scattered transform isFor $x<0$, continuity of the total mass density requires the scattered jump to cancel the incident jump, so $\varphi (b^{+})-\varphi (b^{-})=e^{i{k}_{0}x}$. Its minus transform is $-i/(k+k_0)$. The unknown plate-side jump is the plus function $J^{+}=[{\Phi }^{+}{]}_{b^{-}}^{b^{+}}$. Thus $J=J^{+}-i/(k+k_0)$, and the Wiener-Hopf equation follows:

Use the Wiener-Hopf factorization $L=L^{+}{L}^{-}$, with factors analytic and nonzero in their designated half-planes. Their analytic continuations allocate outgoing modal zeros to $L^{+}$ below the contour, and the opposite zeros to $L^{-}$ above it. Set $C=L^{+}(-k_0)$. Multiplication by $L^{+}$ and pole subtraction giveThe pole in the upper expression is removable by its numerator. The two expressions analytically continue to the common entire function, which is zero under the stipulated edge/growth assumption. Henceand the transformed fields areA constant reciprocal rescaling of the factors does not alter $C{L}^{-}$ or the physical field.

Inside the guide, $\cosh (\gamma y)$ and $\gamma \sinh (\gamma b)$ are even entire functions of $\gamma$, while $L^{-}$ is analytic in the lower half-plane. Thus continuation across the lower branch cut changes none of the interior transform: that cut is removable. The exterior expression retains the cut, corresponding to radiation into the open exterior.

For $x>0$, the inverse-transform contour closes downwards, clockwise. Away from modal cutoffs, its enclosed singularities are the outgoing simple poleswith positive real part for propagating modes and negative imaginary part for decaying modes; $k_0$ is the $n=0$ root. The opposite roots lie above the contour or cancel against zeros of $L^{-}$. Symmetry excludes odd transverse modes. At $n\ge 1$, $\gamma =in\pi /b$, and differentiation of $D(k)=\gamma \sinh (\gamma b)$ givesEach inverse-transform contribution is $-i$ times its residue. Therefore the outgoing modes of an open rigid acoustic waveguide areThe original time factor multiplies this expression. Thus every cut-on mode propagates in the positive $x$ direction; cut-off modes are evanescent waves decaying into the guide, not additional backward waves. Strictly, a complete mode sum includes these evanescent modes as well as propagating ones. The displayed simple-pole amplitudes apply away from exact cutoffs; cutoff values use the outgoing limiting-absorption continuation before taking the limit. The reflected plane-wave amplitude relative to the unit incident wave is $A_0$ above. It has the expected negative sign in the long-wavelength leading kernel approximation $L\simeq b(k^2-k_0^2)$.

Independent metric and Hamiltonian parametrization. The Lorentzian Polyakov action isThe worldsheet metric is independent of the embedding. Its equation sets the traceless worldsheet stress tensor to zero, making the metric locally conformal to the induced worldsheet metric; substitution recovers the Nambu–Goto action. Eliminating momentum in the Nambu-Goto phase-space action instead gives $(\dot{X}-u{X}^{\prime }{)}^2/(2e)-e{T}^2{X}^{\prime 2}/2$. On the positive-lapse branch, choose the independent metric, up to a positive Weyl transformation, asIts determinant is $-e^2{T}^2$. Multiplication by $-T/2$ recovers the reduced density, including the mixed term $-u\dot{X}\cdot X^{\prime }/e$. This proves the phase-space identification of the Polyakov metric.

Conformal gauge and residual coordinate freedom. In the Hamiltonian formulation, conformal gauge is $e=1/T,u=0$. The equations become $\ddot{X}-X^{\prime \prime }=0$, while ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$ remain as constraints. In the metric formulation, conformal gauge is ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$, or simply ${\eta }_{ab}$ after fixing Weyl invariance. The metric parametrization proves these choices equivalent. Gauge-fixing the metric does not discard the Virasoro constraints.

With ${\sigma }^{\pm }=t\pm \sigma$, independent reparameterizations ${\sigma }^{\pm }\mapsto f_{\pm }({\sigma }^{\pm })$ multiply the flat metric by a conformal factor. A compensating Weyl transformation restores its chosen representative. Infinitesimally the Conformal Killing equation is ${\partial }_{-}{c}^{+}=0,{\partial }_{+}{c}^{-}=0$. These are the residual conformal transformations of two-dimensional Minkowski spacetime, with the periodicity conditions appropriate to a closed string.

Ghost determinant and anomaly. For gauge conditions $G_A(\varphi )=0$, the Faddeev-Popov determinant is the determinant of the linearized variationInsert this determinant in the gauge-fixed path integral and exponentiate it with anticommuting ghosts and antighosts: the additional action is proportional to $\int b^A{M}_{AB}{c}^B$. In metric gauge fixing, the trace is removed by Weyl invariance, and a worldsheet diffeomorphism changes the trace-free metric throughConsequently $c^a$ is a vector ghost and $b^{ab}$ a symmetric trace-free antighost. A conventional normalization of the worldsheet ghost action isA rescaling of the antighost changes only this overall kinetic normalization. In conformal gauge, after normalizing the chiral fields, the kinetic terms areThe closed-string ghosts are periodic, with independent left and right systems. Their conformal weights are $2$ for $b$ and $-1$ for $c$, so the supplied bc ghost system expression gives $c_{\mathrm{gh}}=-2(6\cdot 2^2-6\cdot 2+1)=-26$. Each free embedding boson has central charge one per chirality. HenceThis central charge of reparameterization ghosts cancels the string conformal anomaly in each chirality separately. Having two sectors does not double the required target dimension.

Channel variables, poles and the vector state. With particles 1,2 incoming and 3,4 outgoing, the Mandelstam variables are $s=-(p_1+p_2{)}^2$ and $t=-(p_1-p_3{)}^2$. They measure squared center-of-mass energy and momentum transfer. In the specified units ${\alpha }^{\prime }=1$, the external tachyon mass squared is $-1$, so $s+t+u=-4$.

The Gamma function residue at $-n$ is $(-1{)}^n/n!$. At $s=n-1$, its argument $-1-s$ varies with the opposite sign to $s$. The Gamma function recurrence givesThus the Veneziano amplitude pole residues areFor generic fixed $t$ the simple poles are $-1,0,1,\dots$. A zero of the residue product at an exceptional negative integer $t$ makes the corresponding pole removable. Fixed $t$ must be away from its own channel poles; fixing it at a channel singularity does not give an ordinary finite meromorphic function of $s$.