Primitive Dirichlet L Functions and Functional Equations

Write $q=\prod p^{e_p}$. For each $p$, choose the least $c_p\le e_p$ for which the local unit character factors through reduction modulo $p^{c_p}$; the finite set of possible exponents has such a least element.

CRT combines the resulting local characters into a character ${\chi }^{\ast }$ modulo $d=\prod p^{c_p}$, and [F1] shows that its induction has the same values as $\chi$, including zero precisely on the nonunits modulo $q$.

If a primitive character modulo $d^{\mathrm{\prime }}$ induces $\chi$, each of its local exponents must be at least $c_p$; minimality gives $d\mid d^{\mathrm{\prime }}$. If $d^{\mathrm{\prime }}>d$, its local character factors through a proper divisor, contradicting primitivity. Hence $d^{\mathrm{\prime }}=d$ and CRT gives ${\chi }^{\ast }={\chi }^{\ast \mathrm{\prime }}$.

In [F2], the local factors of $\chi$ and ${\chi }^{\ast }$ agree unless $p\mid q$ but $p\nmid d$; at exactly those primes $\chi (p)=0$, while the primitive factor is $(1-{\chi }^{\ast }(p)p^{-s}{)}^{-1}$.

Cancelling all common local factors and multiplying the finite exceptional set gives the displayed identity.

If $(m,q)=1$, substitution $b=am$ permutes the residue classes and gives the sum as $\overline{\chi (m)}\tau (\chi )$.

If $g=(m,q)>1$, choose a unit $u\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}q\mathrm{/}g)$ for which $\chi (u)\mathrm{\ne }1$; otherwise the unit character factors modulo $q\mathrm{/}g$, contrary to [F2]. Multiplication by $u$ leaves $e(am\mathrm{/}q)$ unchanged and multiplies the sum by $\chi (u)$, so the sum is zero.

In the second case $\chi (m)=0$, so the right side is also zero; the two cases prove the formula.

By the definition of the Gauss sum, $\overline{\tau (\chi )}={\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}\overline{\chi (a)}e(-a\mathrm{/}q)\mathrm{.}$ Multiplying by $\tau (\chi )$ and using [F1] with $m=a$ gives $\mathrm{\mid }\tau (\chi ){\mathrm{\mid }}^2={\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}{\sum }_{b<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}\chi (b)e(ab\mathrm{/}q)e(-a\mathrm{/}q)\mathrm{.}$

The inner sum over $a$ is $q$ when $b\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}q)$ and $0$ otherwise. Thus only $b=1$ remains, and the expression in step 1.1 is $q\chi (1)=q$.

Put $G(\xi )={\int }_{\mathbb{R}}{e}^{-\pi x^2}e(-x\xi )dx$. The Gaussian integral The Gaussian integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$, followed by $u=\sqrt{\pi }x$, gives $G(0)=1$. On every compact $\xi$-interval, the derivative of the integrand is dominated by a constant multiple of $\mathrm{\mid }x\mathrm{\mid }{e}^{-\pi x^2}$, so Differentiation under an improper multiple integral under an integrable derivative bound applied to real and imaginary parts gives $G^{\mathrm{\prime }}(\xi )={\int }_{\mathbb{R}}(-2\pi ix)e^{-\pi x^2}e(-x\xi )dx\mathrm{.}$

Since $(e^{-\pi x^2}{)}^{\mathrm{\prime }}=-2\pi x{e}^{-\pi x^2}$, integration by parts on $[-R,R]$ and then $R\to \mathrm{\infty }$ (the boundary term tends to $0$) yields $G^{\mathrm{\prime }}(\xi )=-2\pi \xi G(\xi )$. Hence $(e^{\pi {\xi }^2}G(\xi ){)}^{\mathrm{\prime }}=0$, and step 1.1 gives $G(\xi )=e^{-\pi {\xi }^2}$.

Substitute $u=\sqrt{t}x$ in the defining integral and apply step 2.1 to obtain the stated $t^{-1\mathrm{/}2}$ scaling.

Define the periodization $F(x)={\sum }_{n\in \mathbb{Z}}f(x+n)$. Schwartz decay makes this series, and every termwise derivative series, converge uniformly on compact sets, so $F$ is a smooth $1$-periodic function.

Its $m$th Fourier coefficient is ${\int }_0^{1}F(x)e(-mx)dx={\sum }_{n\in \mathbb{Z}}{\int }_0^{1}f(x+n)e(-m(x+n))dx=\widehat{f}(m),$ where absolute convergence justifies interchange and the intervals $[n,n+1]$ partition $\mathbb{R}$.

The sequence $(\widehat{f}(m){)}_{m\in \mathbb{Z}}$ is rapidly decreasing, so the Fourier series of $F$ converges absolutely and uniformly to $F$. Evaluating at $0$ gives ${\sum }_{n\in \mathbb{Z}}f(n)=F(0)={\sum }_{m\in \mathbb{Z}}\widehat{f}(m)\mathrm{.}$

Decompose the left side by $n\equiv a(\mathrm{m}\mathrm{o}\mathrm{d}q)$ and apply [F1] to each translated, $q$-scaled Schwartz function.

The finite coefficient of $\widehat{f}(m\mathrm{/}q)\mathrm{/}q$ is ${\sum }_{a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q}\chi (a)e(am\mathrm{/}q)$, which is [F2]. Substitution gives the formula.

Suppose first that $\chi$ is nonprincipal, so $q>1$ and $\chi (0)=0$. Put ${\mathrm{\Theta }}_{\chi }(t)={\sum }_{n\in \mathbb{Z}}\chi (n)n^a{e}^{-\pi n^{2}t\mathrm{/}q}$ for $t>0$, with $n^0=1$ also at $n=0$. Since $\chi (-1)=(-1{)}^a$, the positive and negative terms pair. Termwise Mellin integration, absolutely valid for $\mathrm{Re}s>1$, gives ${\int }_0^{\mathrm{\infty }}{\mathrm{\Theta }}_{\chi }(t)t^{(s+a)\mathrm{/}2-1}dt=2{\left(\frac{q}{\pi }\right)}^{(s+a)\mathrm{/}2}\mathrm{\Gamma }((s+a)\mathrm{/}2)L(s,\chi )=2\mathrm{\Lambda }(s,\chi )\mathrm{.}$ The last equality is the exact normalization of Completed primitive Dirichlet L-function.

Apply [F2] to $x^a{e}^{-\pi t{x}^2\mathrm{/}q}$. For $a=0$, [F1] gives its Fourier transform directly. For $a=1$, differentiating that transform with respect to frequency and using $\widehat{xf}(\xi )=-(2\pi i{)}^{-1}(\widehat{f}{)}^{\mathrm{\prime }}(\xi )$ gives the extra factor $-iq\xi \mathrm{/}t$. Thus in both parity cases ${\mathrm{\Theta }}_{\chi }(t)=\frac{\tau (\chi )}{i^a\sqrt{q}}{t}^{-a-1\mathrm{/}2}{\mathrm{\Theta }}_{\bar{\chi }}(1\mathrm{/}t)\mathrm{.}$ The transformed character also has zero constant term because $q>1$.

Split the Mellin integral of step 1.1 at $t=1$ and use step 2.1 on $(0,1)$, then substitute $u=1\mathrm{/}t$. Writing ${ϵ}_{\chi }=\tau (\chi )\mathrm{/}(i^a\sqrt{q})$ gives $2\mathrm{\Lambda }(s,\chi )={\int }_1^{\mathrm{\infty }}{\mathrm{\Theta }}_{\chi }(t)t^{(s+a)\mathrm{/}2-1}dt+{ϵ}_{\chi }{\int }_1^{\mathrm{\infty }}{\mathrm{\Theta }}_{\bar{\chi }}(u)u^{(1-s+a)\mathrm{/}2-1}du\mathrm{.}$ Both theta kernels and all their derivatives in the displayed complex powers decay exponentially as the real variable tends to infinity, since their zero terms vanish and their remaining summands are a fixed polynomial times a Gaussian. Uniform domination on every compact $s$-set makes both integrals entire in $s$. They therefore continue $\mathrm{\Lambda }(s,\chi )$ entirely. Since $1\mathrm{/}\mathrm{\Gamma }((s+a)\mathrm{/}2)$ is entire by [F4] and the other completion factor is entire and nonzero, $L(s,\chi )$ is entire too.

The only primitive principal character has $q=1$ (Primitive Dirichlet characters and conductor). Now $a=0$ and $\mathrm{\Theta }(t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$ has constant term $1$, with $\mathrm{\Theta }(t)=t^{-1\mathrm{/}2}\mathrm{\Theta }(1\mathrm{/}t)$ by [F1] and [F2]. For $\mathrm{Re}s>1$, $2\mathrm{\Lambda }(s)={\int }_0^{\mathrm{\infty }}(\mathrm{\Theta }(t)-1)t^{s\mathrm{/}2-1}dt$. Splitting at $1$ and using $\mathrm{\Theta }(t)-1=t^{-1\mathrm{/}2}(\mathrm{\Theta }(1\mathrm{/}t)-1)+t^{-1\mathrm{/}2}-1$ gives $2\mathrm{\Lambda }(s)={\int }_1^{\mathrm{\infty }}(\mathrm{\Theta }(t)-1)t^{s\mathrm{/}2-1}dt+{\int }_1^{\mathrm{\infty }}(\mathrm{\Theta }(t)-1)t^{(1-s)\mathrm{/}2-1}dt+\frac{2}{s-1}-\frac{2}{s}\mathrm{.}$ The integrals are entire by Gaussian decay. Their boundary terms give exactly simple poles at $1$ and $0$, with nonzero residues $1$ and $-1$ for $\mathrm{\Lambda }$. Multiplication by the entire reciprocal Gamma from [F4] cancels the pole at $0$ without introducing another pole, while its value at $1\mathrm{/}2$ is nonzero by the product formula. Thus $L=\zeta$ has exactly one simple pole, at $1$.

Define, for $t>0$, ${\theta }_{\chi }(t)={\sum }_{n\in \mathbb{Z}}{n}^a\chi (n)e^{-\pi n^{2}t\mathrm{/}q}\mathrm{.}$ For $a=0$, [F3] applied to $x\mapsto e^{-\pi t{x}^2\mathrm{/}q}$ gives ${\theta }_{\chi }(t)=\frac{\tau (\chi )}{\sqrt{q}}{t}^{-1\mathrm{/}2}{\theta }_{\bar{\chi }}(1\mathrm{/}t)\mathrm{.}$ For $a=1$, differentiating the Gaussian transform gives $\widehat{x{e}^{-\pi t{x}^2\mathrm{/}q}}(m\mathrm{/}q)=-im\sqrt{q}{t}^{-3\mathrm{/}2}{e}^{-\pi m^2\mathrm{/}(qt)},$ so [F3] gives the same formula with the extra factor $-i$ and exponent $t^{-3\mathrm{/}2}$. Thus in both cases ${\theta }_{\chi }(t)=\epsilon (\chi )t^{-a-1\mathrm{/}2}{\theta }_{\bar{\chi }}(1\mathrm{/}t)\mathrm{.}$

Suppose first that $q>1$. Then $\chi (0)=0$, so termwise integration for $\mathrm{Re}s>1$ and parity give $2\mathrm{\Lambda }(s,\chi )={\int }_0^{\mathrm{\infty }}{\theta }_{\chi }(t)t^{(s+a)\mathrm{/}2-1}dt\mathrm{.}$ Substitute $t=1\mathrm{/}u$ after using step 1.1; the exponent becomes $u^{(1-s+a)\mathrm{/}2-1}$, so the right side is $2\epsilon (\chi )\mathrm{\Lambda }(1-s,\bar{\chi })$. Thus the displayed equation holds for $q>1$ by [F1].

For $q=1$, put $\theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Step 1.1 becomes $\theta (t)=t^{-1\mathrm{/}2}\theta (1\mathrm{/}t)$, while termwise integration gives $2\mathrm{\Lambda }(s,\mathbf{1})={\int }_0^{\mathrm{\infty }}(\theta (t)-1)t^{s\mathrm{/}2-1}dt(\mathrm{Re}s>1)\mathrm{.}$ Splitting at $1$ and substituting $t=1\mathrm{/}u$ in the first piece yields $2\mathrm{\Lambda }(s,\mathbf{1})={\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{(1-s)\mathrm{/}2-1})dt+\frac{2}{s(s-1)}\mathrm{.}$ The right side is invariant under $s\mapsto 1-s$ and supplies its meromorphic continuation, so $\mathrm{\Lambda }(s,\mathbf{1})=\mathrm{\Lambda }(1-s,\mathbf{1})$. Together with step 2.1 this proves the theorem.

Taking absolute values in [F1] gives $\mathrm{\mid }\epsilon (\chi )\mathrm{\mid }=\mathrm{\mid }\tau (\chi )\mathrm{\mid }\mathrm{/}\sqrt{q}$.

Fact [F2] makes this quotient $1$.

At $s=-a-2m$, $\mathrm{\Gamma }((s+a)\mathrm{/}2)$ has a pole by [F2].

The regularity in [F1] forces $L(s,\chi )$ to cancel that pole. The principal exception is outside the hypothesis.