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This infinite product converges uniformly on compact subsets of the complex plane. Pair the factors at $iz$ and $-iz$ in the Weierstrass product for the reciprocal gamma function, and use the Gamma reflection formula together with $\Gamma (1+w)=w\Gamma (w)$. This gives $[\Gamma (1+iz)\Gamma (1-iz){]}^{-1}=\sinh (\pi z)/(\pi z)$ and the displayed product. The removable value at $z=0$ is one. Matching zeros alone would not exclude multiplication by a nonvanishing entire function; the normalized product fixes that ambiguity.

The Bessel turning-point asymptotic comes from a cubic stationary endpoint, not an ordinary quadratic stationary point. Near $\vartheta =0$,Thus the contributing width is $\vartheta =O(n^{-1/3})$. On writing $\vartheta =(6/n{)}^{1/3}s$, the leading integral isThe oscillatory integral is understood with a vanishing damping factor. Substitution $u=s^3$ and the Gamma function Fourier integral giveThereforeThe equality uses the Gamma reflection formula. Contributions away from the degenerate endpoint are smaller. The $n^{-1/3}$ scale explains why the preceding $n^{-1/2}$ formula cannot be extended directly to zero angle.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute convergence of the integrated series follows from the same expression with $\mathrm{Re}s$ in place of $s$. Integrating termwise and substituting $u=nt$ givesThis proves the Mellin transform representation of the Riemann zeta function.

Take the Hankel contour along the lower side of the negative axis towards zero, around zero counterclockwise, and back along the upper side. For $\mathrm{Re}s>1$ the small-circle contribution tends to zero. On the lower and upper rays, respectively, $\arg t=-\pi ,+\pi$, so the two integrals combine toThe gamma reflection formula now proves agreement of the two representations. At an integer $s=m\ge 2$, the contour integral has a zero while $\Gamma (1-s)$ has a pole: their product is interpreted by its finite limit, equal to $\zeta (m)$, not by multiplying literal zero and infinity.

For nonpositive integers the integrand is single-valued, and the contour closes around zero. Sincethe residue of $t^{-1}/(e^{-t}-1)$ is $-1/2$, giving $\zeta (0)=-1/2$. Alsois odd. Its Laurent series therefore has only odd powers. Multiplication by $t^{-2n-1}$ cannot produce a $t^{-1}$ term, and the subtracted constant cannot do so when $n\ge 1$. Hence the contour integral vanishes and $\zeta (-2n)=0$ for every positive integer $n$.

In a point-particle description of quantum gravity, interactions of a massless spin-two field occur at localized vertices of Feynman diagrams. The gravitational coupling has negative mass dimension for spacetime dimension greater than two, so short-distance loop integrations generate higher-derivative counterterms that cannot all be absorbed into the Einstein action. A string worldsheet instead joins incoming and outgoing strings by a smooth surface. In particular, a closed string splits or joins through a smooth pair-of-pants surface. There is no invariantly distinguished point at which every part of an extended string interacts. The choice of a time slicing can draw a joining point, but it is not a physical pointlike vertex. This removes the point-interaction geometry responsible for arbitrarily localized interactions and introduces the string length ${\ell }_s=\sqrt{{\alpha }^{\prime }}$.

This geometric explanation is supported by the organization of the Polyakov path integral. Introducing a worldsheet metric makes the Nambu–Goto action amenable to gauge fixing. In the Euclidean integral, metrics on a fixed oriented surface are divided by worldsheet diffeomorphisms and Weyl transformations. What remains is a Riemann surface with a conformal structure, together with finitely many worldsheet moduli and the positions of external string vertex operators. The determinants from fixing the metric are represented by Faddeev-Popov ghosts. Schematically, for $n$ closed-string external states,Here ${\mathcal{M}}_{g,n}$ is the moduli space of genus-$g$ punctured surfaces. The dilaton weights a surface by its Euler characteristic $\chi =2-2g$, giving $g_s^{-\chi }$ before the normalization of external states. Every additional handle multiplies the contribution by $g_s^2$. The string genus expansion therefore plays the role of the ordinary loop expansion, with the sphere giving tree level and the torus giving one loop. The critical dimension cancels the worldsheet Weyl anomaly, making this reduction to conformal geometry consistent.

The massless symmetric traceless closed-string state is a graviton, and its long-wavelength interactions reproduce gravity. At distances comparable with ${\ell }_s$, however, an infinite tower of excited states contributes. This is more than a single particle form factor: the same amplitude contains graviton exchange, massive higher-spin exchange and string-scale softness.

The dimensionless variables in the given Virasoro–Shapiro amplitude correspond to $s={\alpha }^{\prime }S/4$, $t={\alpha }^{\prime }T/4$, $u={\alpha }^{\prime }U/4$, where capital letters denote the usual dimensionful Mandelstam invariants. Its constraint $s+t+u=-4$ identifies the external particles as the closed bosonic tachyons, not four gravitons. Nevertheless its intermediate massless pole directly exhibits graviton exchange. The overall normalization will be suppressed below; no additional copy of the amplitude formula is needed.

First, the amplitude is symmetric under permutations of $s,t,u$, expressing crossing symmetry. The poles in any channel areFor generic $t$, the residue can be calculated with the Gamma function pole at $-n$ and its recurrence:The product is $1$ when $n=0$. Indeed, the pole factor contributes $(-1{)}^{n+1}/n!$, while the two remaining Gamma ratios contribute $(-1{)}^n(t+2{)}_n^2/n!$. A residue polynomial of degree $2n$ shows exchange of spins up to $2n$, with leading Regge trajectory $J=2+2s$. At $n=1$, the massless residue is $-(t+2{)}^2$, containing the spin-two exchange. By crossing, the massless $t$-channel pole isfor generic $s$. Thus at large $s$ the pole behaves as $s^2/t$, the characteristic energy dependence of graviton exchange. Its massless propagator produces the long-range gravitational interaction. At low energies the massive poles can instead be expanded into local higher-derivative corrections to the effective gravitational action.

At the string scale one must retain the whole tower. The Gamma reflection formula rewrites the amplitude exactly asAt large $s$ and fixed $t$, away from the resonance poles or with a specified complex continuation, the Gamma ratio scales as $s^{t+1}$. Hence the Regge limit scales as $s^{2+2t}$, up to the signature factor and the displayed $t$-dependent coefficient. In particular, a fixed nonzero momentum transfer changes the point-graviton power law by a factor $s^{2t}$: the graviton belongs to an entire trajectory rather than remaining an isolated spin-two exchange at arbitrarily high energy.

There is stronger suppression when the scattering angle is fixed. Put $t\simeq -as$, $u\simeq -(1-a)s$ with $0<a<1$. Applying Stirling formula to the Gamma functions, after the same treatment of real-axis poles, gives the leading smooth envelopeThe leading $s\log s$ terms cancel and the remaining entropy-like combination is negative. In physical units the exponent is $-{\alpha }^{\prime }Sf(a)/2$. This fixed-angle softness of the closed-string amplitude is absent for an elementary pointlike gravitational vertex. Equivalently, short-distance gravitational scattering probes additional string degrees of freedom. In a four-dimensional long-distance effective setting, a massive exchanged mode contributes a Yukawa-type correction proportional to $e^{-M_{n}r}/r$, with couplings and tensor factors depending on the external states. Such terms are suppressed for $r\gg {\ell }_s$ and become relevant for $r\sim {\ell }_s$. The amplitude establishes the spectrum and short-distance change; it does not by itself give a universal static potential for arbitrary sources. Its bosonic tachyon also signals an unstable vacuum, so it must not be treated as a complete stable model of a gravitational force.

Finally, the modular group prevents overcounting equivalent surfaces and reorganizes the loop ultraviolet region. A torus is $\mathbb{C}/(\mathbb{Z}+\tau \mathbb{Z})$ with $\mathrm{Im}\tau >0$. A change of lattice basis identifiesThus the one-loop integral uses a fundamental domain for $\mathrm{PSL}(2,\mathbb{Z})$, for exampleThe measure $d^2\tau /(\mathrm{Im}\tau {)}^2$ is invariant, and the physical integrand, including matter, ghosts and external insertions, must respect torus modular invariance. In this domain $\mathrm{Im}\tau \ge \sqrt{3}/2$. The region of arbitrarily small proper time that causes ultraviolet divergences in a particle loop is therefore not an independent part of the string integral; a modular transformation relates it to another description already represented in the domain.

This does not remove every possible divergence. The cusp $\mathrm{Im}\tau \to \infty$ is a long tube, and factorization through it can produce infrared divergences, massless tadpoles or the bosonic tachyon divergence. At higher genus, degenerating cycles similarly factorize into propagating intermediate string states. The essential improvement is the relation between worldsheet geometry, the full spectrum and the quotient by large diffeomorphisms; infrared and unstable-background problems remain separate questions. A short-distance expansion that keeps only the graviton discards precisely this structure.

The frequency-independent measure factor cancels in the ratio:Upon analytic continuation in frequency, this has poles at ${\omega }_1=0,\pm 2\pi ir/\beta$ and zeros at ${\omega }_2=0,\pm 2\pi ir/\beta$, for $1\le r\le N$, apart from cancellations when numerator and denominator vanish together. They match those of $\sinh (\beta {\omega }_2/2)/\sinh (\beta {\omega }_1/2)$. To justify equality, rather than merely matching this divisor, use the hyperbolic-sine infinite productIt follows from pairing the Weierstrass product for the reciprocal gamma function at opposite imaginary arguments and using the Gamma reflection formula. Normal convergence on compact sets follows from $\sum r^{-2}<\infty$. Thus, for positive real frequencies,The cutoff ratio determines the frequency dependence, agreeing with the thermal partition function of a quantum harmonic oscillator.

The printed assertion about an absolute ${\lim }_N{\mathcal{Z}}_N$ needs a normalization qualification: independence of $A_N$ from $\omega$ alone does not ensure existence of a nonzero limit. Set $B_N=A_N{\prod }_{r=1}^N{\nu }_r^{-2}$. Then ${\mathcal{Z}}_N=B_N{\omega }^{-1}{\prod }_r(1+{\omega }^2/{\nu }_r^2{)}^{-1}$, so if $B_N\to B\ne 0$, its limit is $B\beta /[2\sinh (\beta \omega /2)]$. The choice $B_N=1/\beta$ gives the required limit. Conversely, $B_N=(1+(-1{)}^N/2)/\beta$ is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing entire functions; the convergent normalized product is what rules them out here.

For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute integrability, bounded near zero by a constant times $t^{\mathrm{Re}s-2}$, permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius $0<r<2\pi$, and return above the cut. Take $-\pi <\arg t<\pi$. The circle excludes every nonzero pole $2\pi i\mathbb{Z}$ of the denominator. For $\mathrm{Re}s>1$ its radius can shrink to zero, and the lower and upper rays give respectively $-e^{-i\pi s}\Gamma (s){\zeta }_R(s)$ and $e^{i\pi s}\Gamma (s){\zeta }_R(s)$. By the Gamma reflection formula,away from integer $s$ in that half-plane, with limiting equality at the removable points.

With the circle radius kept fixed, the contour integral is an entire function of $s$: the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by $\Gamma (1-s)$ is analytic for $\mathrm{Re}s<1$ and meromorphic globally. The apparent singularities at positive integers $s\ge 2$ are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only $s=1$ remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on $\mathbb{C}\setminus \{1\}$. Keeping the circular segment is essential when the integral is not locally integrable at zero.

At $s=-1$ the two rays cancel, since $t^{-2}$ is single-valued. The Laurent expansionshows that the residue of $t^{-2}/(e^{-t}-1)$ is $-1/12$. Since $\Gamma (2)=1$, the residue theorem gives

The given Euler product for the gamma function has the finite approximantsaway from the poles. Split the denominator of ${\Gamma }_{2m+1}(2z)$ into even and odd factors. The resulting exact identity isThe right side does not depend on $z$, so its limit is a constant $C$. Evaluate the ratio at $z=1/2$: $\Gamma (1)=1$, and the gamma reflection formula, together with positivity on the positive real axis, gives $\Gamma (1/2)=\sqrt{\pi }$. HenceThis is the gamma duplication formula. The argument first applies where the factors are finite; the identity then holds as an identity of meromorphic functions, with values at poles interpreted accordingly.

With $u=x^4$, the change of variables formula givesThe beta--gamma identity, $\Gamma (1/2)=\sqrt{\pi }$, and the Gamma reflection formula giveTherefore

Take the Hankel contour about the negative real axis: it starts at $-\infty$ just below the cut, circles the origin counterclockwise, and returns to $-\infty$ just above the cut. The notation $(0+)$ records this return side of the branch point. Choose the branch $-\pi <\arg t<\pi$ for $t^{z-1}$.

For $\mathrm{Re}z>1$, collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula $\Gamma (z)\Gamma (1-z)=\pi /\sin (\pi z)$, recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on $z$. The apparent poles of $\Gamma (1-z)$ are cancelled by zeros of the contour integral except at $z=1$. It therefore gives the analytic continuation of the Riemann zeta function to $\mathbb{C}\setminus \{1\}$.

Use the Hankel contour about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi <\arg t<\pi$. For $\mathrm{Re}s>0$, the small circle vanishes and the two banks giveThe gamma reflection formula then gives

Near zero,Subtracting any desired number of these terms makes the local contour integral converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma (1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in \mathbb{C}$.