= Solution
Locally in the space of <worldsheet metrics>, divide out the volume of <worldsheet diffeomorphisms> and <Weyl transformations>. With stabilizer zero modes omitted, the inverse <Faddeev-Popov determinant> at a metric $g$ is
$$
\boxed{\Delta_{\rm FP}^{-1}[g]=\int D\omega\,Dv\ \delta\!\left[2\omega g_{ab}+\nabla_a v_b+\nabla_b v_a\right].}
$$
This is the linearized orbit integral: the delta functional is the product of the metric-component delta functions, and its Jacobian is the inverse determinant of the gauge-orbit map. The omitted zero modes and any metric moduli must be handled separately; this local formula is not an integration over the physical moduli themselves.
The <trace elimination in string Faddeev-Popov gauge fixing> is essential. In two dimensions, split the variation as
$$
\delta g_{ab}=(2\omega+\nabla\cdot v)g_{ab}+(P_1v)_{ab},\qquad (P_1v)_{ab}=\nabla_av_b+\nabla_bv_a-g_{ab}\nabla\cdot v.
$$
The second term is trace-free. Integration over $\omega$ removes the trace delta function, leaving $\int Dv\,\delta[P_1v]=(\det{}'P_1)^{-1}$, with the measure understood through the metric inner products. Equivalently, a Fourier representation of the metric delta functional introduces a commuting symmetric tensor $\beta^{ab}$. Integration over $\omega$ enforces $g_{ab}\beta^{ab}=0$, and the remaining exponent is proportional to $\int\sqrt g\,\beta^{ab}(P_1v)_{ab}$. This gives the same inverse <Faddeev-Popov determinant> and explains why its tensor variable must be trace-free.
For the <Polyakov path integral>, insert the gauge-fixing identity
$$
1=\Delta_{\rm FP}[g]\int D\omega\,Dv\ \delta[g^{\omega,v}-\widehat g(m)],
$$
where $\widehat g(m)$ is a representative depending on the <worldsheet moduli> $m$. Gauge invariance of the action, measure, and <Faddeev-Popov determinant> lets us change variables along the gauge orbit. Its volume cancels the original gauge-group denominator, leaving an integral over $X$, the moduli, and the determinant evaluated at $\widehat g$. This is how the determinant enters the numerator of the gauge-fixed integral, despite the inverse determinant occurring in the orbit integral above.
A <fermionic Gaussian integral> exponentiates $\det{}'P_1$ using an anticommuting vector $c^a$ and an anticommuting symmetric trace-free tensor $b^{ab}$:
$$
\det{}'P_1\ \propto\ \int Db\,Dc\ \exp\!\left[-C\int d^2\sigma\sqrt g\,b^{ab}(P_1c)_{ab}\right].
$$
The constant $C$ can be absorbed into $b$. In local complex <conformal gauge>, trace-freeness sets $b_{z\bar z}=0$. The only surviving components of the <worldsheet diffeomorphism ghost operator> are proportional to $\bar\partial c^z$ and $\partial c^{\bar z}$. Rescaling the antighost therefore gives exactly the chiral <worldsheet ghost action>
$$
\boxed{S_{\rm gh}=\int d^2z\left(b_{zz}\bar\partial c^z+b_{\bar z\bar z}\partial c^{\bar z}\right).}
$$
The chiral <bc ghost system> has weights $h_b=2$, $h_c=-1$, with $b(z)c(w)\sim1/(z-w)$. Its <worldsheet stress tensor> is $T_{\rm gh}=-2:b\partial c:-:(\partial b)c:$. Fermionic double <Wick contractions> in $T_{\rm gh}T_{\rm gh}$ give the fourth-order coefficient $-13$, hence <central charge> $c_{\rm gh}=-26$. More generally this calculation gives $c=1-3(2h_b-1)^2$. Each embedding scalar contributes one, so the total chiral <central charge> is $D-26$. The gauge-fixed quantum theory must have no <worldsheet Weyl anomaly>; consequently \b[the <critical dimension of the bosonic string> is $D=26$]. Zero-mode insertions for <Conformal Killing vector fields> and moduli do not change this local anomaly calculation.
For the <Veneziano amplitude>, first examine an ordered beta-function term. A pole occurs when $-\alpha's-1=-n$, with $n=0,1,2,\ldots$, so the exchanged masses follow the <open bosonic string mass spectrum>
$$
\boxed{M_n^2=\frac{n-1}{\alpha'},\qquad n=0,1,2,\ldots.}
$$
There is a tachyon at level zero, a massless vector at level one, then an infinite tower with equally spaced squared masses. In oscillator quantization these levels are $N=n$; the poles alone do not determine all oscillator degeneracies or all states that might fail to couple to the chosen external particles.
The <Veneziano amplitude pole residue> gives considerably more information than a pole position. Near $s_n=(n-1)/\alpha'$, use $\Gamma(-n+\epsilon)\sim(-1)^n/(n!\epsilon)$ and $\epsilon=-\alpha'(s-s_n)$. The finite <gamma function> ratio is a product of $n$ linear factors. Thus, for generic $t$,
$$
\operatorname*{Res}_{s=s_n}B(-\alpha's-1,-\alpha't-1)=-\frac1{\alpha'n!}\prod_{j=1}^{n}(\alpha't+j+1).
$$
The empty product at $n=0$ is one. This degree-$n$ angular polynomial decomposes into spin <partial waves>; it reveals the spins that couple, up to maximal spin $n$ in the ordered channel, and the corresponding products of three-point couplings. For example, the level-one linear residue encodes vector exchange. A single external-state amplitude need not separate all states with the same mass and spin.
There is a qualification specific to the sum printed in the question. For four identical on-shell tachyons, $s+t+u=-4/\alpha'$. At $\alpha's=n-1$, put $P_n(t)=\prod_{j=1}^n(\alpha't+j+1)$. Then $\alpha'u=-n-3-\alpha't$, and reversing the product gives $P_n(u)=(-1)^nP_n(t)$. The $t,u$ beta-function term is regular at a generic $s$-channel pole. The residue of the full symmetric amplitude is therefore proportional to
$$
-\frac{1+(-1)^n}{\alpha'n!}\,P_n(t).
$$
This is <odd-level pole cancellation in the symmetric Veneziano amplitude>. \b[The fully symmetrized identical-tachyon amplitude has only even-level poles at generic angle], including the tachyon but not the massless vector. This is an external-state selection rule, not removal of odd-level states from the string theory. Ordered amplitudes, or suitable amplitudes with color factors and different external states, retain those exchanges. Thus interpreting every candidate <gamma function> pole as a pole of the printed sum would give a false inference about that amplitude.
At high energy with $x=-t/s\in(0,1)$ fixed, the <gamma function> reflection formula followed by the <Stirling formula> gives the <fixed-angle softness of an open-string amplitude>:
$$
\log|A|=-\alpha's\left[-x\log x-(1-x)\log(1-x)\right]+O(\log s),
$$
up to oscillatory factors and possible additional cancellations. The bracket is positive at a genuinely nonforward fixed angle. This is the envelope away from resonance poles, or the corresponding analytically continued asymptotic regime; it is not a uniform bound through real-axis poles. \b[String scattering is exponentially soft at fixed angle], unlike the typical power-law behavior of point-particle interactions. The finite string length distributes a hard collision over an extended object. This is evidence of improved ultraviolet behavior, not by itself a proof that every loop amplitude is finite. The fixed-angle limit also differs from the fixed-$t$ <Regge limit>.
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