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A meromorphic continuation extends a holomorphic function from a connected open set to a larger connected domain as a meromorphic function, allowing isolated poles. Equality on the initial open set determines the extension uniquely by the identity theorem. Integral formulas whose integrands depend holomorphically on a parameter often supply such an extension after contour deformation or subtraction of singular terms.

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 $\xi (s)=\xi (1-s)$.

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