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Tate's local theory at the infinite places

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).

At an infinite place, a Gaussian replaces the characteristic function of the integral ring. A polynomial attached to the angular character then selects the correct Gamma factor. Fourier transformation turns that polynomial into its opposite angular type, with a phase that records the degree. We will compute this phase as well as prove the functional equation for every Schwartz function.

We retain the quotient convention of Tate's local theory at the finite places: the Fourier-transformed integral is the factor times the original integral. The local characters are those classified in Quasi-characters and Hecke characters. For a real place write \[ c_{\epsilon,s}(x)=\operatorname{sgn}(x)^\epsilon |x|^s, \qquad \epsilon\in\{0,1\},\quad s\in\mathbb C; \tag{1} \] for a complex place write \[ c_{n,s}(z)=(z/|z|)^n |z|_{\mathbb C}^s, \qquad n\in\mathbb Z,\quad |z|_{\mathbb C}=z\bar z. \tag{2} \] Here the unadorned \(|z|\) in the angular part is the ordinary radius. In both cases \(\sigma(c)=\operatorname{Re}s\), and \(\check c=c^{-1}|\cdot|_F\). Thus the dual parameters are \((\epsilon,1-s)\) over \(\mathbb R\) and \((-n,1-s)\) over \(\mathbb C\).

Analysis dependency details: supplementary exposition by GPT-6 Astra (OpenAI), Ultra, 11 October 2026, CC0.

A common normalization

For \(F=\mathbb R\) or \(\mathbb C\), define \[ \widehat f(y)=\int_F f(x)\psi_F(xy)\,dx, \qquad Z(f,c)=\int_{F^\times}f(x)c(x)\,d^\times x. \tag{3} \] The choices are fixed in the following table. Write \(z=u+iv\) at a complex place.

| Object | Real place | Complex place | |---|---|---| | Normalized norm | \(\lvert x\rvert\) | \(\lvert z\rvert_{\mathbb C}=z\bar z\) | | Additive character | \(e^{-2\pi i x}\) | \(e^{-2\pi i(z+\bar z)}\) | | Self-dual additive measure \(dx\) | \(dx\), ordinary Lebesgue measure | \(2\,du\,dv\) | | Multiplicative measure | \(dx/\lvert x\rvert\) | \(dx/(\pi\lvert z\rvert_{\mathbb C})\) | | Gamma factor | \(\Gamma_{\mathbb R}(s)=\pi^{-s/2}\Gamma(s/2)\) | \(\Gamma_{\mathbb C}(s)=2(2\pi)^{-s}\Gamma(s)\) |

The complex additive pairing is bilinear multiplication, not the Hermitian pairing: if \(y=a+ib\), its kernel is \(e^{-4\pi i(ua-vb)}\). The measure \(2\,du\,dv\) and the reversal in the second coordinate give the inversion formula \(\widehat{\widehat f}(x)=f(-x)\). These additive Fourier conventions and inversion are established in Propositions 5.1–5.2 of Additive characters, self-dual measures and Poisson summation on the adèles.

In polar coordinates, the chosen complex multiplicative measure is \[ d^\times z=4\,\frac{dr}{r}\frac{d\theta}{2\pi} =2\,\frac{dt}{t}\frac{d\theta}{2\pi},\qquad t=r^2. \tag{4} \] Thus the angular group has mass two when the norm coordinate has measure \(dt/t\). The alternative \(dx/(2\pi|z|_{\mathbb C})\) has half this mass, and gives half our Gaussian zeta integral. It gives the same \(\rho\) and epsilon factor, because a multiplicative measure constant cancels in their defining ratio. This explicit choice accounts for the factor two in \(\Gamma_{\mathbb C}\).

The space \(\mathcal S(F)\) consists of smooth functions on the underlying real vector space for which every derivative, multiplied by every polynomial, is bounded. Differentiating a Fourier integral inserts polynomial factors; integrating by parts transfers powers of the Fourier variable to derivatives. All these integrals are absolutely convergent for a Schwartz function. Consequently \(\widehat f\in\mathcal S(F)\). For example the bound for \(y^\alpha\partial_y^\beta\widehat f\) follows from the \(L^1\)-norm of the corresponding derivative of \(x^\beta f\), with fixed powers of \(2\pi\) or \(4\pi\). This also justifies the differential identities used below.

Exact substitution proofs

Polar coordinates and scalar substitutions. The complete planar formula is proved in Banach foundation bridges, Section 15.6, (PC1)–(PC7). For every nonnegative Lebesgue measurable \(h\), and for every absolutely integrable complex \(h\), it gives \[ \int_{\mathbb R^2}h(u,v)\,du\,dv =\int_0^{2\pi}\int_0^\infty h(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. \tag{S1} \] The map \((r,\theta)\mapsto(r\cos\theta,r\sin\theta)\) is a Borel bijection from \((0,\infty)\times[0,2\pi)\) onto the punctured plane; its determinant is \(r\). The cited proof obtains sector areas from the linear Jacobian, proves equality of the resulting measures, and treats completed measurability, the collapsed origin and the duplicated angular endpoint. Thus it is not an appeal to an unproved nonlinear substitution theorem. Applied to the absolute value, (S1) also gives the exact equivalence of absolute integrability. Integrals of completed-measurable functions have the usual almost-everywhere section interpretation; the functions used in this lesson are Borel.

Our multiplicative measure is \(2\,du\,dv/(\pi r^2)\). Consequently (S1) gives \[ \frac{2}{\pi r^2}\,r\,dr\,d\theta =4\,\frac{dr}{r}\frac{d\theta}{2\pi}. \] This verifies the first equality of (4) with its original factor four. For the second equality use \(t=r^2\), so \(dt=2r\,dr\) and \(dr/r=dt/(2t)\).

The scalar substitutions here are justified on compact subintervals by Real-variable calculations, Lemma 1.3, which proves substitution from the fundamental theorem and chain rule. Its endpoint argument uses monotone convergence for nonnegative integrands and dominated convergence for absolutely convergent complex integrals. In particular \(t=r^2\) and \(t=\pi r^2\) are first used on \(0<a\le r\le b<\infty\), then \(a\downarrow0\), \(b\uparrow\infty\). No inverse derivative at radius zero is required. The same proof covers the affine substitution \(t=(1+a)/2\) and the substitution \(b=a^2\) on \(0<a<1\) in (32). Their endpoint integrals are absolute when \(\operatorname{Re}z>0\): after squaring, the absolute-value bound is \(b^{-1/2}(1-b)^{\operatorname{Re}z-1}\), integrable at both endpoints. All positive-base complex powers use the real logarithm. For the Mellin integrals the already stated half-plane bounds and Schwartz decay supply the required endpoint integrability.

The beta substitution, with its full domain. Define \[ \begin{aligned} \Psi:(0,\infty)\times(0,1)&\longrightarrow(0,\infty)^2, &\Psi(r,t)&=(rt,r(1-t)),\\ \Psi^{-1}(x,y)&=(x+y,x/(x+y)), &D\Psi(r,t)&=\begin{pmatrix}t&r\\1-t&-r\end{pmatrix}. \end{aligned} \] Both compositions are the identity on the displayed domains; in particular every denominator is positive. The determinant is \(-rt-r(1-t)=-r\), and its absolute value is \(r\). For every nonnegative Borel \(H\), Tonelli and the affine substitution formula, Section 15.1, (LP4)–(LP5), give \[ \begin{aligned} \int_0^\infty\!\int_0^\infty H(x,y)\,dy\,dx &=\int_0^\infty\!\int_x^\infty H(x,r-x)\,dr\,dx\\ &=\int_0^\infty\!\int_0^r H(x,r-x)\,dx\,dr\\ &=\int_0^\infty\!\int_0^1 H(rt,r(1-t))\,r\,dt\,dr. \end{aligned} \tag{S2} \] The first step is the translation \(y=r-x\) at fixed \(x>0\). The second applies Tonelli to the Borel function \(\mathbf1_{\{0<x<r\}}H(x,r-x)\), defined to be zero outside that region. The third uses \(x=rt\) at fixed \(r>0\), retaining its scale factor \(r\). These nonnegative identities allow infinite values and require no prior integrability assumption.

There is also no restriction to the uncompleted Borel measure. Apply (S2) to Borel indicators. Since \(\Psi\) and its inverse are continuous on their domains, this identifies the image of the measure \(r\,dr\,dt\) with plane Lebesgue measure. On the increasing sets \([1/k,k]\times[1/k,1-1/k]\), \(k\ge3\), the weight \(r\) is bounded above and below by positive constants. Thus weighted null sets and coordinate null sets coincide. The Borel bijection therefore transports Borel null sets in both directions; subsets of these null sets give the same assertion for the completions. Indicator approximation proves (S2) for every nonnegative Lebesgue measurable \(H\). Applying it to \(|H|\) gives absolute integrability on one side if and only if on the other. Applying it to the positive and negative parts of the real and imaginary components proves (S2) for absolutely integrable complex \(H\), with almost-everywhere section conventions.

For (31), take \(H(x,y)=e^{-x-y}x^{u-1}y^{v-1}\), where \(\operatorname{Re}u,\operatorname{Re}v>0\). Its absolute integral is the product of the two convergent positive Euler integrals with exponents \(\operatorname{Re}u\) and \(\operatorname{Re}v\). Formula (S2) transforms it into \[ e^{-r}r^{u+v-1}t^{u-1}(1-t)^{v-1}\,dt\,dr. \] The powers multiply without branch ambiguity because \(r,t,1-t\) are positive. Absolute Fubini now gives \(\Gamma(u)\Gamma(v)=\Gamma(u+v)B(u,v)\). The beta factor converges at zero and one precisely under the displayed real-part hypotheses. The previously proved nonvanishing of Gamma permits division by \(\Gamma(u+v)\). This proves the exact substitution used in Proposition 8.4 for complex parameters, not only for positive real ones.

Mellin continuation without a functional equation

The measure-theoretic inputs are dominated convergence, Theorem 2.2, and sigma-finite Radon Tonelli and Fubini, Theorem 4.4. Lebesgue measure on \(\mathbb R\) and \(\mathbb R^2\), and their restrictions to the radial intervals used below, are sigma-finite: bounded intervals or boxes form countable finite-measure covers. The positive and absolutely integrable versions of that theorem are the ones used here; no unrestricted Fubini assertion for a non-sigma-finite Haar measure is needed.

For parameter differentiation, Lemma 1.4 of Real-variable calculations proves the real-parameter assertion from an integrable derivative bound. The complex version used here follows by the same difference-quotient argument. Let each \(H(s,\cdot)\) be measurable, let \(H(s_0,\cdot)\) be integrable at the centre of a small closed parameter disc, and let \(H(\cdot,x)\) be holomorphic on a neighbourhood of that disc. Suppose that \(|\partial_sH(s,x)|\le M(x)\) throughout the disc with \(M\) integrable. The fundamental theorem along a segment first gives \(|H(s,x)-H(s_0,x)|\le |s-s_0|M(x)\), so every kernel in the disc is integrable. It also gives \[ \frac{H(s+h,x)-H(s,x)}h =\int_0^1\partial_sH(s+th,x)\,dt. \] For an interior parameter, take \(h\) small enough that its segment stays in the disc. The quotient is measurable, its absolute value is at most \(M(x)\), and it tends to \(\partial_sH(s,x)\) for every complex sequence \(h\to0\). The derivative is measurable as such a limit. Dominated convergence proves complex differentiation under the integral. For the Mellin kernels, choose a slightly larger compact parameter disc inside the stated convergence half-plane. At zero its lower real-part bound gives a power of \(r\); at infinity its upper bound and Schwartz decay give an integrable negative power. The resulting bounds absorb each fixed power of \(|\log r|\): after \(r=e^{-u}\) at zero or \(r=e^u\) at infinity, this is boundedness of \(u^k e^{-\delta u}\) for \(\delta>0\). It follows, for example, from the exponential series, which bounds that expression by \(k!\delta^{-k}\). The same argument applies to every derivative order separately.

First define \(\Gamma(w)=\int_0^\infty e^{-t}t^{w-1}\,dt\) for \(\operatorname{Re}w>0\). Integration by parts gives \(\Gamma(w+1)=w\Gamma(w)\), so this continues meromorphically, with simple poles at \(0,-1,-2,\ldots\). We will also use its absence of zeros; the following elementary argument fixes precisely that fact.

Repeated integration by parts gives \[ \int_0^1 t^{w-1}(1-t)^Ndt =\frac{N!}{w(w+1)\cdots(w+N)}. \tag{5} \] After \(u=Nt\), the left side multiplied by \(N^w\) tends to \(\Gamma(w)\). Indeed \((1-u/N)^N\leq e^{-u}\) on \([0,N]\), and dominated convergence applies to \(u^{w-1}\), uniformly on compact subsets of \(\operatorname{Re}w>0\). For the scaled integrals, the finite reciprocal identity and locally uniform product proof below justify the limit without assuming nonvanishing; they give \[ \frac1{\Gamma(w)} =w e^{\gamma w}\prod_{k=1}^\infty(1+w/k)e^{-w/k}, \qquad \gamma=\lim_{N\to\infty} \left(\sum_{k=1}^N\frac1k-\log N\right). \tag{6} \] The harmonic-sum limit exists by comparison with the integral of \(1/t\). On a fixed compact set, all sufficiently large factors have logarithm \(O(k^{-2})\), uniformly there. Thus the product defines an entire function and is nonzero away from the displayed linear-factor zeros. At \(0,-1,-2,\ldots\) exactly one such factor vanishes simply. Equality initially holds in the right half-plane and then agrees with the reciprocal of the recurrence continuation. In particular \(\Gamma\) has no zeros and has exactly the asserted simple poles.

Why the reciprocal limit does not assume nonvanishing. For integers \(N\ge1\) and \(\operatorname{Re}w>0\), define \[ \begin{aligned} I_N(w)&=N^w\int_0^1t^{w-1}(1-t)^N\,dt,\\ P_N(w)&=wN^{-w}\prod_{k=1}^N(1+w/k),\\ I_N(w)P_N(w)&=1. \end{aligned} \tag{G1} \] To see the finite identity (5) with its endpoint conditions, put \(J_N(w)=\int_0^1t^{w-1}(1-t)^Ndt\). For \(N\ge1\), integrate the derivative of \(t^w/w\) against \((1-t)^N\). Both boundary terms vanish when \(\operatorname{Re}w>0\), and \(J_N(w)=N J_{N-1}(w+1)/w\). The terminal integral is \(J_0(w+N)=1/(w+N)\). Iteration gives exactly (5). The last equality in (G1) follows from it, since \(w\prod_{k=1}^N(1+w/k)=w(w+1)\cdots(w+N)/N!\). Here \(N^{\pm w}=\exp(\pm w\log N)\) uses the real logarithm. None of the finite denominators in (5) vanishes in the right half-plane. The product \(P_N\), unlike \(I_N\), is defined and entire on all of \(\mathbb C\).

Write \(H_N=\sum_{k=1}^N1/k\) and \(\delta_N=H_N-\log N\). The integral comparison is explicit: \[ \begin{aligned} \delta_N&>0,\\ 0<\delta_N-\delta_{N+1} &=\int_N^{N+1}\frac{dx}{x}-\frac1{N+1} <\frac1N-\frac1{N+1}. \end{aligned} \tag{G2} \] Indeed \(H_N\ge\int_1^{N+1}dx/x>\log N\). The displayed monotone, bounded sequence therefore has a limit \(\gamma\), and summing the last bound from \(N\) onwards gives \(0\le\delta_N-\gamma\le1/N\).

To justify the infinite product, fix \(R>0\) and an integer \(K>2R\). For \(k>K\), on the disc \(|w|<k\), put \[ L_k(w)=\sum_{j=2}^{\infty}\frac{(-1)^{j+1}}j(w/k)^j. \] The power-series lemma and the locally uniform holomorphic-limit theorem, Lemma 3.1 and Corollary 3.5 of the Cauchy reading, apply to these exact scalar series. Termwise differentiation gives \[ L_k'(w)=\frac1{k+w}-\frac1k. \] Consequently the derivative of \((1+w/k)\exp(-w/k-L_k(w))\) is zero on that disc; its value at zero is one. A zero derivative makes it constant by integration along line segments. Hence \[ \exp L_k(w)=(1+w/k)e^{-w/k}. \] This constructs the logarithm needed for the tail without choosing a logarithm across any zero of a finite factor. On \(|w|\le R\), \[ |L_k(w)|\le\sum_{j=2}^{\infty}(R/k)^j \le \frac{2R^2}{k^2},\qquad \sum_{k>M}\frac1{k^2}\le\frac1M. \tag{G3} \] The last inequality follows by comparison with \(\int_M^\infty x^{-2}dx\). Thus the logarithmic tail converges uniformly on every smaller compact disc and defines a holomorphic sum. Exponentiation is compatible with that uniform convergence, since the sums are uniformly bounded on each compact disc.

For \(N\ge K\), the exact finite identity \[ P_N(w)=w e^{\delta_N w} \prod_{k=1}^{K}(1+w/k)e^{-w/k} \exp\!\left(\sum_{k=K+1}^{N}L_k(w)\right) \] therefore converges uniformly on \(|w|\le R\) to \[ F(w)=w e^{\gamma w} \prod_{k=1}^{K}(1+w/k)e^{-w/k} \exp\!\left(\sum_{k>K}L_k(w)\right). \tag{G4} \] Different choices of \(R,K\) agree because they are limits of the same \(P_N\). Corollary 3.5 consequently makes \(F\) entire. Its exponential factors never vanish, since \(e^ze^{-z}=1\). In a neighbourhood of any fixed point, choose \(K\) larger than twice its modulus and then a small disc about the point. Formula (G4) shows that the only zeros there are those of \(w\prod_{k=1}^K(1+w/k)\). They are the distinct points \(0,-1,\ldots,-K\), each of multiplicity one. Varying the disc proves that \(F\) has precisely the simple zeros \(0,-1,-2,\ldots\), and no others.

The Euler-integral limit also holds locally uniformly, with an explicit majorant. If a compact set of parameters lies in \(\sigma_0\le\operatorname{Re}w\le\sigma_1\), where \(0<\sigma_0\le\sigma_1\), write \[ I_N(w)=\int_0^\infty u^{w-1}a_N(u)\,du,\qquad a_N(u)= \begin{cases}(1-u/N)^N,&0\le u<N,\\0,&u\ge N.\end{cases} \] For \(0\le u<N\), integrating \(1/(1-v)\ge1\) gives \(\log(1-u/N)\le-u/N\), so \(0\le a_N(u)\le e^{-u}\). For each fixed \(u\), the same logarithmic series gives \(N\log(1-u/N)=-u+O(u^2/N)\) as \(N\to\infty\); hence \(a_N(u)\to e^{-u}\). The error, uniformly over the parameter compact set, is bounded by the integrable function \[ g(u)= \begin{cases} u^{\sigma_0-1}e^{-u},&0<u\le1,\\ u^{\sigma_1-1}e^{-u},&u>1. \end{cases} \] At zero its integral is bounded by \(1/\sigma_0\). At infinity an exponential-series bound makes every fixed power times \(e^{-u/2}\) bounded, leaving an integrable multiple of \(e^{-u/2}\). Apply dominated convergence, Theorem 2.2 of the integration reading, to \(\max(u^{\sigma_0-1},u^{\sigma_1-1})|a_N(u)-e^{-u}|\). Its integral tends to zero and bounds \(\sup_w|I_N(w)-\Gamma(w)|\) on that compact set. Thus \(I_N\to\Gamma\) locally uniformly there.

Taking limits in the finite identity (G1) now gives \(\Gamma(w)F(w)=1\) on \(\operatorname{Re}w>0\). This proves nonvanishing there before any limiting reciprocal is taken. The meromorphic function \(1/F\) extends this integral, and agrees with the recurrence continuation because they agree on the initial half-plane. Equivalently, for each integer \(m\ge1\), on \(\operatorname{Re}w>-m\) away from \(0,-1,\ldots,-(m-1)\), the recurrence writes that continuation as \[ \Gamma(w)=\frac{\Gamma(w+m)}{w(w+1)\cdots(w+m-1)}; \] these expressions agree on their overlaps by the original recurrence, and the meromorphic identity principle proved later in this lesson identifies them with \(1/F\). The simple zeros of \(F\) give exactly the asserted simple poles of \(\Gamma\); a reciprocal of an entire function has no zeros at its finite holomorphic points.

The residues used in the exercises follow with their signs unchanged. Apply the recurrence \(m+1\) times, and cancel only the explicit denominator factor \(w+m\). Since \(\Gamma(1)=\int_0^\infty e^{-t}dt=1\), for every integer \(m\ge0\), \[ \lim_{w\to-m}(w+m)\Gamma(w) =\frac{\Gamma(1)}{\prod_{j=0}^{m-1}(-m+j)} =\frac{(-1)^m}{m!}. \tag{G5} \] For \(m=0\) the product is empty and equals one. The numerator is holomorphic near \(1\); the other denominator factors are nonzero near \(-m\), so this limit is the residue of a simple pole, not a cancellation of unexamined singularities.

The Taylor remainder used below is uniform in direction. To see this directly, let \(f\) be \(C^{M+1}\) on a neighbourhood of a closed ball about zero in \(\mathbb R^d\), where \(d=1\) or \(2\), and set \(g(t)=f(tx)\) for \(0\le t\le1\). The fundamental theorem and chain rule, Lemma 0.1(2), give \[ g(1)=\sum_{j=0}^M\frac{g^{(j)}(0)}{j!} +\frac1{M!}\int_0^1(1-t)^M g^{(M+1)}(t)\,dt. \] Indeed the integral remainder for \(M=0\) is \(g(1)-g(0)\). Integration by parts for \(M\ge1\) makes the remainder equal to the preceding remainder minus \(g^{(M)}(0)/M!\); induction proves the formula. Repeated differentiation of \(f(tx)\) gives \[ g^{(j)}(t)=\sum_{|\alpha|=j}\frac{j!}{\alpha!} x^\alpha\partial^\alpha f(tx). \] If every derivative of total order \(M+1\) is bounded in absolute value by \(B\) on that ball, the remainder is at most \[ \frac{B}{(M+1)!}\left(\sum_{j=1}^d|x_j|\right)^{M+1}. \] Thus it is \(O(|x|^{M+1})\), uniformly over all directions. This proves the real expansion and the uniform circular remainder used in (9) and (11), including their use for complex-valued smooth functions.

Mellin lemma. For every \(f\in\mathcal S(F)\), \(Z(f,c)\) is absolutely convergent for \(\operatorname{Re}s>0\), locally uniformly there. It has meromorphic continuation to all \(s\). Its possible poles are simple, at \[ s=-\epsilon-2j\quad(F=\mathbb R),\qquad s=-|n|/2-j\quad(F=\mathbb C),\qquad j\geq0. \tag{7} \] Moreover \(Z(f,c_{\epsilon,s})/\Gamma_{\mathbb R}(s+\epsilon)\) and \(Z(f,c_{n,s})/\Gamma_{\mathbb C}(s+|n|/2)\) are entire.

Proof. Over \(\mathbb R\), split the integral into its two signs: \[ Z(f,c_{\epsilon,s})=\int_0^\infty [f(r)+(-1)^\epsilon f(-r)]r^{s-1}\,dr. \tag{8} \] Boundedness near zero and rapid decay at infinity prove convergence. For a compact set of parameters, the same bounds work with the smallest real part at zero and the largest at infinity. They also absorb every power of \(\log r\), proving holomorphy under differentiation in \(s\).

The integral on \([1,\infty)\) is entire by rapid decay. On \((0,1)\), Taylor expansion to arbitrary order \(M\) gives \[ f(r)+(-1)^\epsilon f(-r) =2\!\sum_{\substack{0\leq k\leq M\\ k\equiv\epsilon\ (2)}} \frac{f^{(k)}(0)}{k!}r^k+O(r^{M+1}). \tag{9} \] Each displayed term integrates to a constant divided by \(s+k\). The remainder is holomorphic for \(\operatorname{Re}s>-M-1\). Increasing \(M\) gives the continuation and its poles in (7). All are among the simple poles of \(\Gamma_{\mathbb R}(s+\epsilon)\).

Over \(\mathbb C\), put \[ P_n f(r)=\int_0^{2\pi}f(re^{i\theta})e^{in\theta} \frac{d\theta}{2\pi}. \tag{10} \] For \(\operatorname{Re}s>0\), polar coordinates and absolute Fubini give \(Z(f,c_{n,s})=4\int_0^\infty P_n f(r)r^{2s-1}\,dr\). The real Taylor polynomial of \(f(u+iv)\) can equally be written as a sum of monomials \(z^a\bar z^b\). Angular integration retains exactly those with \(a-b=-n\). Their total degrees are \(|n|+2j\). The Taylor remainder is uniformly \(O(r^{M+1})\) on the circle. Hence \[ P_n f(r)=\sum_{|n|+2j\leq M}b_jr^{|n|+2j} +O(r^{M+1}). \tag{11} \] Integration on \((0,1)\) produces denominators \(2s+|n|+2j\); the remainder converges for \(2\operatorname{Re}s+M+1>0\). Infinity again gives an entire integral. This proves the complex assertion. Finally (6) shows that division by either Gamma factor cancels the possible simple poles and introduces no new ones. The resulting functions are entire. ∎

Absolute convergence and angular cancellation

The angular projection controls the poles and can improve convergence of the projected radial expression. It does not by itself improve absolute convergence of the original integral (3), because the angular character has modulus one. The identities obtained by first integrating over angles or signs hold initially in the common half-plane \(\operatorname{Re}s>0\). Beyond it, a convergent projected expression represents the continuation and need not be an absolutely convergent original integral.

For example, take \(f(z)=e^{-2\pi|z|^2}\) and \(n=1\). Then \(P_1f(r)=0\) for every \(r\), so its projected radial integral is zero for every \(s\). At \(s=0\), however, the original absolute integral is

\[ \int_{\mathbb C^\times}|f(z)c_{1,0}(z)|\,d^\times z =4\int_0^\infty e^{-2\pi r^2}\frac{dr}{r}=\infty. \]

On \((0,1)\) the Gaussian is at least \(e^{-2\pi}\), giving logarithmic divergence. More generally, this original integral converges exactly for \(\operatorname{Re}s>0\), independently of \(n\). Over \(\mathbb R\), the even Gaussian and \(\epsilon=1\) give the same distinction: \(f(r)-f(-r)=0\), but its absolute integral is \(2\int_0^\infty e^{-\pi r^2}r^{\operatorname{Re}s-1}\,dr\).

The original absolute domain extends left when the test itself vanishes at zero. Suppose its first nonzero Taylor degree is \(k\). Over \(\mathbb R\), with leading coefficient \(a_k\ne0\), Taylor's theorem and \(\bigl||a+b|-|a|\bigr|\leq|b|\) give

\[ |f(r)|+|f(-r)|=2|a_k|r^k+O(r^{k+1}). \]

Thus the original integral converges at zero exactly when \(\operatorname{Re}s>-k\). Over \(\mathbb C\), write the leading real homogeneous Taylor polynomial as \(P_k\). The Taylor remainder is uniform on the unit circle, so

\[ \int_0^{2\pi}|f(re^{i\theta})|\frac{d\theta}{2\pi} =r^k\bigl(A+O(r)\bigr),\qquad A=\int_0^{2\pi}|P_k(e^{i\theta})|\frac{d\theta}{2\pi}>0. \]

The positivity follows because a nonzero homogeneous polynomial is nonzero somewhere on the unit circle and hence on an open arc. With the measure \(4\,dr/r\,d\theta/(2\pi)\) of (4), integrability is therefore exactly \(2\operatorname{Re}s+k>0\). At either boundary the leading term produces logarithmic divergence. Rapid Schwartz decay handles infinity for every fixed parameter and uniformly on compact parameter sets.

If \(f\) is flat at zero, Taylor's remainder is \(O(r^M)\) for every \(M\). The original integral is then absolutely convergent for every \(s\); these bounds also absorb logarithmic powers, so differentiation under the integral proves it entire. The common half-plane in the Mellin lemma is sufficient for every Schwartz test. The larger domains for the polynomial Gaussian tests below come from their actual polynomial vanishing, not merely angular cancellation.

Tests that detect the angular type

Define the real tests \[ g_0(x)=e^{-\pi x^2},\qquad g_1(x)=xe^{-\pi x^2}, \tag{12} \] and, for \(N=|n|\), the complex tests \[ h_n(z)= \begin{cases} \bar z^n e^{-2\pi z\bar z},&n\geq0,\\ z^{-n}e^{-2\pi z\bar z},&n<0. \end{cases} \tag{13} \] The exponents here are nonnegative integers. In particular the second case is a polynomial, not a singular function.

Proposition 8.2. For the measures in the table, \[ Z(g_\epsilon,c_{\epsilon,s})=\Gamma_{\mathbb R}(s+\epsilon), \qquad \widehat g_\epsilon=(-i)^\epsilon g_\epsilon, \tag{14} \] and \[ Z(h_n,c_{n,s})=\Gamma_{\mathbb C}(s+N/2), \qquad \widehat h_n=(-i)^N h_{-n}. \tag{15} \] The integral formulas converge respectively for \(\operatorname{Re}s>-\epsilon\) and \(\operatorname{Re}s>-N/2\), and then hold meromorphically everywhere.

Proof. The character's sign cancels the polynomial's sign in (12). The real integral is therefore \[ 2\int_0^\infty e^{-\pi r^2}r^{s+\epsilon-1}\,dr =\pi^{-(s+\epsilon)/2}\Gamma((s+\epsilon)/2), \tag{16} \] by \(t=\pi r^2\). At a complex place the angular factors in \(h_n c_{n,s}\) cancel. Equation (4) gives \[ 4\int_0^\infty e^{-2\pi r^2}r^{2s+N-1}\,dr =2(2\pi)^{-(s+N/2)}\Gamma(s+N/2). \tag{17} \] These substitutions also give the exact convergence conditions.

The Gaussian Fourier transforms are already computed with these measures in Proposition 5.2. For the real polynomial, differentiation under the integral gives \[ \widehat{xf}(y)=-\frac1{2\pi i}\frac{d}{dy}\widehat f(y). \tag{18} \] Applying this to \(g_0\) gives \(\widehat g_1=-iyg_0(y)=-ig_1(y)\).

For the complex calculation let \(w=a+ib\) and \(G(w)=e^{-2\pi w\bar w}\). Direct differentiation of the bilinear kernel gives \[ \widehat{\bar z f} =-\frac1{4\pi i}(\partial_a+i\partial_b)\widehat f, \qquad \widehat{zf} =-\frac1{4\pi i}(\partial_a-i\partial_b)\widehat f. \tag{19} \] Put \(D_+=\partial_a+i\partial_b\). Then \(D_+w=0\) and \(D_+G=-4\pi wG\). Induction consequently gives \(D_+^N G=(-4\pi)^N w^N G\), with no additional lower-degree terms. Since \(\widehat G=G\), (19) gives \(\widehat{\bar z^N G}=(-i)^Nw^NG\). Similarly \(D_-\bar w=0\) and \(D_-G=-4\pi\bar wG\), so \(\widehat{z^NG}=(-i)^N\bar w^NG\). These are exactly (15), for both signs of \(n\). ∎

The normalization of a test function and that of the multiplicative measure should be compared together. With the complex measure \(dx/|z|_{\mathbb C}\) used by Poonen and Leahy, the same unscaled polynomial Gaussian \(h_n\) has zeta integral \(\pi\Gamma_{\mathbb C}(s+|n|/2)\), since that measure is \(\pi\) times (3). Poonen’s \(\pi^{-1}h_n\) therefore gives our \(\Gamma_{\mathbb C}\); Leahy’s \((2\pi)^{-1}h_n\), equation (4.10), gives half of it. These scalar changes multiply both sides of a local zeta-integral identity equally. The phases for our negative basic character are fixed by the differentiated Fourier transforms (14)–(15).

The functional equation for all tests

Theorem 8.1. For \(0<\operatorname{Re}s<1\) and \(f,g\in\mathcal S(F)\), \[ Z(f,c)Z(\widehat g,\check c) =Z(g,c)Z(\widehat f,\check c). \tag{20} \] There is a nonzero meromorphic function \(\rho(c,\psi_F,dx)\), independent of \(f\), such that \[ Z(\widehat f,\check c)=\rho(c,\psi_F,dx)Z(f,c) \tag{21} \] as meromorphic functions of \(s\), for every Schwartz function \(f\).

Proof. Write \(d^\times x=\kappa dx/|x|_F\), where \(\kappa=1\) over \(\mathbb R\) and \(\kappa=1/\pi\) over \(\mathbb C\). The Mellin lemma and preservation of the Schwartz space show that all four factors in (20) converge absolutely in the strip. Thus their first product is \[ \kappa^2\int_{F^\times}\int_{F^\times} f(x)\widehat g(y)c(x)c(y)^{-1}\frac{dx}{|x|_F}\,dy. \tag{22} \] Change \(y=xz\). Additive measure scales by \(|x|_F\), so this equals \[ \kappa^2\int_{F^\times}c(z)^{-1} \left(\int_F f(x)\widehat g(zx)dx\right)dz. \tag{23} \] The absolute-value double integral is finite by undoing that substitution in (22). For each fixed \(z\), the integral of \(|f(x)g(t)|\) is finite, since both functions are integrable on the additive space. Therefore Fubini at this fixed parameter gives \[ \int_F f(x)\widehat g(zx)dx =\int_F\widehat f(zt)g(t)dt. \tag{24} \] Insert (24) in (23). The absolute-value double integral of the new expression is finite as well: reversing the change \(y=zt\) gives the product of the two convergent absolute zeta integrals for \(g\) and \(\widehat f\). That same substitution now evaluates it as the right side of (20). A singleton has additive measure zero over either field, so the inserted or removed points at zero do not affect these steps.

Take \(g=g_\epsilon\) or \(h_n\). Its zeta integral is the nonzero Gamma function from Proposition 8.2; it has neither poles nor zeros in the strip. Dividing (20) by this integral proves (21) there. The Mellin lemma continues all its integrals, and the explicit Gamma ratio from (14) or (15) continues \(\rho\). The holomorphic identity theorem, Theorem 3.7, gives the meromorphic version needed here as follows. For the difference of the two meromorphic sides, the set of points where its meromorphic germ is zero is open. At any limit point \(a\) of this set, take a small disc on which multiplication by \((s-a)^m\), for some nonnegative integer \(m\), removes the possible pole. The product is holomorphic and vanishes on a nonempty open subset of that disc, so the holomorphic identity theorem makes it zero on the whole disc. The set of zero germs is therefore closed as well. It contains the strip, and the plane is connected, so it is the whole plane. This proves (21) meromorphically, including at every pole. Fourier inversion applied twice also gives \(\rho(c)\rho(\check c)=c(-1)\), so the factor is not identically zero. ∎

The two-variable proof has the same algebra as Theorem 7.2, but its convergence here comes from Schwartz bounds. The proof justifies the fixed-parameter Fourier interchange separately from the two absolutely convergent double integrals; it requires no absolutely convergent oscillatory triple integral.

Gamma factors and epsilon phases

Proposition 8.3. Define \(L_\infty\) and epsilon using the same quotient direction as at finite places: \[ \rho(c)=\epsilon(c)\frac{L_\infty(\check c)}{L_\infty(c)}. \tag{25} \] For our additive characters and self-dual measures their values are \[ \begin{array}{c|c|c|c} F&c&L_\infty(c)&\epsilon(c)\\\hline \mathbb R&\operatorname{sgn}^{\epsilon}|\cdot|^s &\Gamma_{\mathbb R}(s+\epsilon)&(-i)^\epsilon\\ \mathbb C&(z/|z|)^n|z|_{\mathbb C}^s &\Gamma_{\mathbb C}(s+|n|/2)&(-i)^{|n|}. \end{array} \tag{26} \] In particular \[ \rho(c_{\epsilon,s})=(-i)^\epsilon \frac{\Gamma_{\mathbb R}(1-s+\epsilon)} {\Gamma_{\mathbb R}(s+\epsilon)},\qquad \rho(c_{n,s})=(-i)^{|n|} \frac{\Gamma_{\mathbb C}(1-s+|n|/2)} {\Gamma_{\mathbb C}(s+|n|/2)}. \tag{27} \] These are meromorphic identities; at poles one uses continuation rather than ratios of divergent integrals.

Proof. Evaluate (21) on the tests of Proposition 8.2. The real Fourier transform has the scalar \((-i)^\epsilon\). The complex transform has scalar \((-i)^{|n|}\) and the opposite angular polynomial, which is precisely the test for \(\check c=c_{-n,1-s}\). Equations (14)–(15) consequently give (27), hence (26). The Mellin lemma shows that these \(L\)-factors remove every possible pole of a test integral; the selected tests have normalized integral one. Thus the factors account for the full family of local integrals, rather than for the selected Gaussians alone. ∎

For example, the characters with \(n=1\) and \(n=-1\) both have epsilon \(-i\); those with \(n=2\) and \(n=-2\) both have epsilon \(-1\). Replacing \(|n|\) by \(n\) in the phase would fail for negative odd \(n\). The dual-product check is \[ \epsilon(c)\epsilon(\check c)= \begin{cases} (-1)^\epsilon,&F=\mathbb R,\\ (-1)^{|n|},&F=\mathbb C, \end{cases} =c(-1), \tag{28} \] in agreement with Fourier inversion. Each epsilon has absolute value one for these basic characters, for every \(s\); in particular it has absolute value one for a unitary character times \(|\cdot|_F^{1/2}\).

The dependence on the additive data has a useful uniform form. If \(dx\) is multiplied by \(a>0\), Fourier transformation and epsilon are multiplied by \(a\). If \(\psi_F\) is replaced by \(\psi_F(rx)\) while \(dx\) is held fixed, the change of variable \(y\mapsto ry\) in the dual zeta integral gives \[ \epsilon(c,\psi_{F,r},dx) =c(r)|r|_F^{-1}\epsilon(c,\psi_F,dx). \tag{29} \] The measure that remains self-dual for this new character is \(|r|_F^{1/2}dx\), so the combined factor is \(c(r)|r|_F^{-1/2}\). These statements follow from (3) and (21), exactly as in Proposition 7.4.

In particular replacing the negative character in our table by the positive one is \(r=-1\), with no change of self-dual measure. It multiplies epsilon by \((-1)^\epsilon\) or \((-1)^n\), producing \(i^\epsilon\) and \(i^{|n|}\). This is the convention in [Deligne 1973, (3.4.1)–(3.4.2)]; its displayed characters are inverse powers of the real or complex embedding, whose angular parameters give the same parity or absolute degree. The difference in phase is fully accounted for by the additive character.

Duplication and the two complex embeddings

Proposition 8.4. As meromorphic functions, \[ \Gamma_{\mathbb C}(s) =\Gamma_{\mathbb R}(s)\Gamma_{\mathbb R}(s+1). \tag{30} \]

Proof. For \(\operatorname{Re}u,\operatorname{Re}v>0\), change variables \((x,y)=(rt,r(1-t))\) in the absolutely convergent product of two Euler Gamma integrals. The absolute Jacobian is \(r\); the beta-substitution proof above establishes the change of variables and absolute Fubini for these complex parameters. This proves \[ B(u,v):=\int_0^1t^{u-1}(1-t)^{v-1}dt =\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}. \tag{31} \] For \(\operatorname{Re}z>0\), the substitution \(t=(1+a)/2\), followed by symmetry and \(b=a^2\), gives \[ B(z,z)=2^{1-2z}\int_{-1}^1(1-a^2)^{z-1}da =2^{1-2z}B(1/2,z). \tag{32} \] Use (31), cancel the nonzero \(\Gamma(z)\), and rearrange: \[ \Gamma(z)\Gamma(z+1/2) =2^{1-2z}\Gamma(1/2)\Gamma(2z). \tag{33} \] The real Gaussian integral in Proposition 5.2, together with \(t=\pi r^2\), gives \(\Gamma(1/2)=\sqrt\pi\). Put \(z=s/2\) in (33) and multiply by \(\pi^{-s-1/2}\). Its right side becomes \(2^{1-s}\pi^{-s}\Gamma(s)=2(2\pi)^{-s}\Gamma(s)\), while its left side becomes \(\Gamma_{\mathbb R}(s)\Gamma_{\mathbb R}(s+1)\). This proves (30) in a half-plane; meromorphic continuation proves it everywhere. ∎

The shift by one is essential. The complex Gamma factor is a product of an even real factor and an odd real factor, both with the same variable \(s\). It is not the square of the even factor. The identity also explains why the two real types together naturally supply the conventional factor two at a complex place.

Examples and exercises

For the trivial character of \(\mathbb R^\times\), the Gaussian gives \(L_\infty(s)=\pi^{-s/2}\Gamma(s/2)\) and epsilon one. This is the factor in the completed Riemann zeta function. At each finite rational place the standard integral-ring test gives \((1-p^{-s})^{-1}\), by Proposition 7.1. Thus the product test supplies \(\Gamma_{\mathbb R}(s)\prod_p(1-p^{-s})^{-1}\) in \(\operatorname{Re}s>1\). Its global continuation and functional equation require the global argument of the next lesson; they do not follow from a single local factor.

For a Gaussian-integer Hecke character whose actual complex component is \((z/|z|)^4\), Proposition 8.3 gives \(\Gamma_{\mathbb C}(s+2)\) and epsilon \((-i)^4=1\). The ideal-type inverse discussed in Theorem 6.3 may instead give angular exponent \(-4\). Its Gamma factor and epsilon are identical, since both depend on the absolute degree, although its actual character and dual are different.

  1. Easy. Evaluate \(\int_{\mathbb R^\times}e^{-\pi x^2}|x|^s\,dx/|x|\), giving its domain of absolute convergence and all its poles after continuation.
  2. Medium. Compute the Fourier transform of \(xe^{-\pi x^2}\) with the negative real additive character. Also state the transform for the positive character and verify the Fourier-square identity on this odd test.
  3. Medium. Prove \(\Gamma_{\mathbb C}(s)=\Gamma_{\mathbb R}(s)\Gamma_{\mathbb R}(s+1)\) from the Euler integrals, including every constant.
  4. Hard. Compute the complex epsilon factor for every integer angular exponent \(n\), including negative \(n\). Verify its dual-product identity. Determine its change for \(\psi_{\mathbb C}(rz)\), first with the old additive measure and then with the new self-dual measure; explain the effect of halving \(d^\times z\).

Solutions

Solution 1. Splitting the signs gives \(2\int_0^\infty e^{-\pi r^2}r^{s-1}dr\). With \(t=\pi r^2\), one has \(dr/r=dt/(2t)\), so the result is \(\pi^{-s/2}\Gamma(s/2)\). At zero absolute integrability is exactly \(\operatorname{Re}s>0\); the Gaussian handles infinity for every parameter. The recurrence and product (6) show simple poles precisely at \(s=0,-2,-4,\ldots\). At \(s=-2j\) their residues are \(2(-1)^j\pi^j/j!\): the residue of \(\Gamma(w)\) at \(-j\) is \((-1)^j/j!\), and the change \(w=s/2\) contributes the factor two. There are no other poles.

Solution 2. The transform of the even Gaussian is itself. Differentiating its Fourier integral gives \(\widehat{xf}=-(2\pi i)^{-1}(\widehat f)'\). Hence \(\widehat{xe^{-\pi x^2}}=-iy e^{-\pi y^2}\). Changing to the positive character replaces the Fourier variable by \(-y\), giving \(+iy e^{-\pi y^2}\). Under the negative convention the transform applied twice multiplies this test by \((-i)^2=-1\), equal to its value at \(-x\). Under the positive convention the multiplier is \(i^2=-1\) as well. The real odd epsilon is correspondingly \(-i\) or \(+i\), not one.

Solution 3. The product \(\Gamma(u)\Gamma(v)\) is absolutely integrable for positive real parts. The change \((x,y)=(rt,r(1-t))\), with Jacobian \(r\), gives (31). For \(u=v=z\), substitute \(t=(1+a)/2\) in the beta integral: the prefactor is \(2^{1-2z}\). Integrating the even function on \([-1,1]\) and then putting \(b=a^2\) yields \(B(z,z)=2^{1-2z}B(1/2,z)\). Applying (31) and canceling \(\Gamma(z)\) gives \(\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt\pi\Gamma(2z)\). Now \(z=s/2\) and the factors \(\pi^{-s/2}\pi^{-(s+1)/2}\) give \(2^{1-s}\pi^{-s}\Gamma(s)=\Gamma_{\mathbb C}(s)\). All cancellations take place in a right half-plane, where (6) guarantees nonvanishing; meromorphic continuation establishes the identity for every parameter.

Solution 4. If \(n\geq0\), choose \(h_n=\bar z^nG\). Formula (19), with \(D_+w=0\), gives \(\widehat h_n=(-i)^n w^nG\). The original integral is \(\Gamma_{\mathbb C}(s+n/2)\), and the transformed test integrated against \(c_{-n,1-s}\) gives \((-i)^n\Gamma_{\mathbb C}(1-s+n/2)\). Their ratio proves epsilon \((-i)^n\). If \(n<0\), set \(N=-n\) and choose \(h_n=z^NG\). The operator \(D_-\) gives \(\widehat h_n=(-i)^N\bar w^NG\), and the same division proves epsilon \((-i)^N\). Thus the answer is \((-i)^{|n|}\) in all cases.

The dual has exponent \(-n\), so the epsilon product is \((-i)^{2|n|}=(-1)^n=c_{n,s}(-1)\), as required. Write \(r=Re^{i\alpha}\), with \(R>0\). Holding the old measure \(2\,du\,dv\) fixed, (29) multiplies epsilon by \(e^{in\alpha}R^{2s-2}\). Renormalizing the additive measure to be self-dual multiplies it by an additional \(|r|_{\mathbb C}^{1/2}=R\). The combined factor is therefore \(e^{in\alpha}R^{2s-1}\). If \(s=1/2+it\), this has absolute value one, providing the central normalization check. Halving the multiplicative measure halves both sides of (21) and changes neither rho nor epsilon. With that alternative measure the Gaussian integral is \(\Gamma_{\mathbb C}(s+|n|/2)/2\); doubling the test would restore normalized integral one.

Prerequisites and further directions

References

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