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Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

Use a genus-zero Polyakov path integral, target signature $(-,+,\dots ,+)$ and all four momenta incoming. Define the Mandelstam variables by $s=-(p_1+p_2{)}^2$, $t=-(p_1+p_3{)}^2$, $u=-(p_1+p_4{)}^2$. The string mass-shell condition is $p_i^2=4/{\alpha }^{\prime }$ and momentum conservation is ${\sum }_i{p}_i=0$, soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.

The free-boson worldsheet propagator is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give $g_s^4$ and the sphere contributes $g_s^{-2}$ through the string genus expansion, leaving $g_s^2$.

The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take $(z_1,z_2,z_3,z_4)=(z,0,1,\infty )$. The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply $|z_{23}{z}_{24}{z}_{34}{|}^2$. At finite $z_4$, this determinant grows as $|z_4{|}^4$, while the matter correlator decays as $|z_4{|}^{-4}$ by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight-$(1,1)$ matter operator at infinity is normalized with $|z_4{|}^4$, while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure $d^{2}z=idz\wedge d\bar{z}=2dxdy$. With $dxdy$ instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed $2\pi$ prefactor.

Set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$. Then $a+b+c=1$ and the integrand is $|z{|}^{2a-2}|1-z{|}^{2b-2}$. To evaluate the complex beta integral first work where $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c>0$. This ensures convergence near $0$, $1$ and infinity. Set $v=1-a$ and $w=1-b$. Schwinger parameterization givesCompleting the square, the $dxdy$ Gaussian integral is $\pi (\lambda +\rho {)}^{-1}\exp [-\lambda \rho /(\lambda +\rho )]$. Change variables to $q=\lambda +\rho$ and $x=\lambda /q$; their Jacobian determinant is $q$. Integrating $q$ gives $\Gamma (v+w-1)[x(1-x){]}^{-(v+w-1)}$. The remaining integral is the ordinary beta function $B(1-w,1-v)=B(b,a)$. ThusRestoring $d^{2}z=2dxdy$ and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.

The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal $1/\Gamma$ is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting $s,t,u$ leaves it unchanged. Its generic channel poles areand the same tower in $t$ and $u$. They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.

A precise check of factorization is the Virasoro–Shapiro amplitude pole residue. Near $a=-n$, set $c=1+n-b$ in the nonsingular factor. The gamma function recurrence givesCombining this with $\Gamma (a)\sim (-1{)}^n/[n!(a+n)]$ showsThe residue is a degree-$2n$ polynomial in $t$, consistent with exchange up to spin $2n$ and scattering-amplitude factorization. At exceptional kinematics reciprocal Gamma zeros can remove apparent poles. In particular, when both $a$ and $b$ approach nonpositive integers, the zero from $1/\Gamma (a+b)$ cancels the putative double pole, leaving channel simple-pole terms. Thus one must not count products of numerator poles without using $a+b+c=1$.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

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.

The Mellin transform iswhere this integral converges. Rapid decay at infinity makes the integral over $[1,\infty )$ an entire function of $s$, but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for $\mathrm{Re}s>-{\sigma }_1$.

For $\mathrm{Re}s>1$, expand $(e^{\pi y}-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-\pi ry}$. Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral

There is a missing hypothesis in the general continuation claim: one needs ${\sigma }_j\to +\infty$. The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract $r$ terms of the asymptotic expansion:Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is holomorphic on $\mathrm{Re}s>-{\sigma }_{r+1}$: on every compact subset, its integrand and all its $s$ derivatives are dominated by an integrable power of $y$ times a power of $|\log y|$. The expressions for successive $r$ agree on their common initial domain and hence everywhere they overlap by the identity theorem. As ${\sigma }_{r+1}\to \infty$, these half-planes cover $\mathbb{C}$. Taking $r\ge j$ with ${\sigma }_{r+1}>{\sigma }_j$ isolates the term $c_j/(s+{\sigma }_j)$, while all other terms are holomorphic near $-{\sigma }_j$. Thus the corrected claim isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.

To show why the correction matters, set ${\sigma }_j=1-1/j$, $c_j=2^{-j}$, andwhere $\eta$ is continuous, equals one for $0<y\le 1$ and vanishes for $y\ge 2$. On $[0,1]$, the remainder after division by $y^{{\sigma }_{r+1}}$ is a uniformly convergent series of nonnegative powers of $y$, so it extends continuously to zero with value $c_{r+1}$. Away from zero, define $g_r$ by the required remainder quotient; it is continuous on the rest of $[0,\infty )$ as well. Thus all the printed hypotheses hold. Its Mellin transform equalsThe series continues meromorphically on $\mathrm{Re}s>-1$ and has genuine poles at $-1+1/j$, accumulating at $-1$. A meromorphic function on $\mathbb{C}$ cannot have such an accumulation of poles. This is the accumulating asymptotic exponents obstruct Mellin continuation counterexample.

For the Gamma function, the Gamma function recurrence givesAt $s=-j$ the numerator is $\Gamma (1)=1$ and the product of the nonzero denominator factors is $(-1{)}^{j}j!$. Therefore the residues of the Gamma function are

The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\ge 1$, the residue of its transform is $B_n{\pi }^{n-1}/n!$, while the residue of ${\pi }^{-s}\Gamma (s)$ is ${\pi }^{n-1}(-1{)}^{n-1}/(n-1)!$. Their quotient defines a holomorphic continuation of the Riemann zeta function at this point, including when the first residue is zero. Dividing gives the Bernoulli formula for zeta values at nonpositive integers:In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta (0)=-1/2$.

In the integral defining the Beta function, put $t=(1+x)/2$ and then $u=x^2$. Symmetry givesThis proves the identity for $\mathrm{Re}z>0$. Both sides extend by analytic continuation to the same meromorphic function, equivalently by the beta--gamma identity and the Gamma duplication formula. The identity holds meromorphically for every $z\in \mathbb{C}$; both sides are finite for $z\notin \{0,-1,-2,\dots \}$ and have matching poles at the excluded points.

If ${\check{H}}^1(S,\mathcal{O})=0$, that cocycle is a Čech coboundary: there are $f_i\in \mathcal{O}(U_i)$ withSet $F=-f_0$ on $U_0$ and $F=P_m-f_m$ on $U_m$. These formulas agree on overlaps, so the sheaf gluing axiom gives a global meromorphic function. Moreover $F-P_m=-f_m$ is holomorphic near $x_m$. Thus $F$ solves the Mittag-Leffler problem on a Riemann surface with precisely the prescribed principal parts.

The exact sequenceinduces a long exact sequence in sheaf cohomology. Under $H^0({\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast })=\mathrm{Div}(X)$ and $H^1({\mathcal{O}}_X^{\ast })=\mathrm{Pic}(X)$, its connecting homomorphism is the divisor-to-Picard map $D\mapsto \mathcal{O}(D)$. Exactness identifies its kernel with divisors of global nonzero meromorphic functions, namely principal divisors:

For $\Re s>1$, the absolutely convergent Dirichlet seriesdefines the Riemann zeta function. To continue it, use partial summation in Stieltjes form:Since $0\le \{x\}<1$, the final integral converges locally uniformly for $\Re s>0$ and is holomorphic there. The displayed expression is consequently a meromorphic function on that half-plane, with its only pole at $s=1$. Since $s/(s-1)$ has residue one there, so does $\zeta$. Agreement in $\Re s>1$ makes this continuation unique by the identity theorem.

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

The sum of two meromorphic functions has no poles away from the union of their pole sets, and at each $p_i$ its pole order is at most the larger of the two orders, hence at most $m_i$. Scalar multiplication cannot increase a pole order. The zero function belongs to the set, so $V$ is a complex vector space.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

A modular function of weight $k$ and level $\Gamma (1)$ is a meromorphic function $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma (1)$, and having a meromorphic Fourier expansion at the cusp infinity. Equivalently, $f{|}_k\gamma =f$ for the slash operator for modular forms.

A modular form is a modular function that is holomorphic on $\mathfrak{h}$ and holomorphic at infinity. Its Fourier expansion therefore has the formSince $-I\in \Gamma (1)$, a nonzero level-one form must have even weight.

The inverse is nonconstant; locally at zero,If the assumed meromorphic function $\mathcal{H}$ had no poles, it would be entire. It is bounded on the compact closure of a fundamental parallelogram, and its two periods then make it bounded throughout the complex plane. The Liouville theorem would force it to be constant, a contradiction. Therefore $\mathcal{H}$ has at least one pole, as stated by the general result that a nonconstant elliptic function has a pole.

Choose holomorphic coordinates centered at $p$. Locally, the blowup of a complex manifold at a point iswith $\sigma (z,[\ell ])=z$; away from $p$ this is an isomorphism, so it glues to $X\setminus \{p\}$. The exceptional divisor is $E={\sigma }^{-1}(p)\cong {\mathbb{CP}}^{n-1}$.

The proper transform is the closure of ${\sigma }^{-1}(Y\setminus \{p\})$. If $p\notin Y$, it is isomorphic to $Y$. If $p\in Y$, the holomorphic implicit function theorem supplies coordinates in which $Y=\{z_1=0\}$. In the blowup chart with $z_i=t$ and $z_j=t{u}_j$ for $j\ne i$, every chart with $i\ne 1$ describes the proper transform by $u_1=0$, while the chart $i=1$ does not meet it. These are smooth coordinate hypersurfaces, so $\tilde{Y}$ is smooth.

For a divisor $D$, the line bundle associated to a divisor $\mathcal{O}(D)$ consists locally of meromorphic functions $f$ such that $(f)+D\ge 0$. Pulling back a local defining function for $Y$ shows that its divisor iswhere $m$ is the order of vanishing at $p$ of a local defining function for $Y$. ThereforeBecause $Y$ is smooth, $m=1$ when $p\in Y$ and $m=0$ when $p\notin Y$.

Applying the definition with $D=E$ gives directlyThe map sends a section to its local meromorphic coefficient relative to the canonical meromorphic section of $\mathcal{O}(E)$; the divisor inequality is exactly the condition that these coefficients define a holomorphic section, and the inverse construction is local multiplication by that canonical section.

At a point $x$ of a Riemann surface, the principal part of a meromorphic function is the negative-degree part of its Laurent series in a local coordinate. It is equivalently a germ in the quotient ${\mathcal{K}}_x/{\mathcal{O}}_x$.

The residue of a meromorphic function $f$ at an isolated singularity $a$ is the coefficient of $(z-a{)}^{-1}$ in its Laurent series. At a simple pole,

The sheaf ${\mathcal{K}}_X$ assigns to an open subset of a complex manifold its meromorphic functions. Its subsheaf ${\mathcal{K}}_X^{\ast }$ consists of the nonzero meromorphic functions, while ${\mathcal{O}}_X^{\ast }$ consists of the nowhere-zero holomorphic functions.

A nonconstant meromorphic function on a compact Riemann surface with exactly one simple pole gives a degree-one holomorphic map to the sphere. A degree-one map is a biholomorphism. The surface therefore has genus zero and the function is a global rational coordinate.