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.

Mellin-transform ${\sum }_{n\ge 1}\chi (n)n^a{e}^{-\pi n^{2}t\mathrm{/}q}$; for $\mathrm{Re}s>1$ this equals a nonzero constant times $\mathrm{\Lambda }(s,\chi )$.

Apply [F1] in [F2] to transform the theta kernel under $t\mapsto 1\mathrm{/}t$. Splitting the Mellin integral at $1$ gives two rapidly decaying integrals, hence an entire continuation when $\chi$ is nonprincipal.

The only primitive principal case is $q=1$. Splitting the Mellin integral for the Jacobi theta kernel and separating its constant term continues ${\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2)\zeta (s)$ meromorphically, with the two boundary terms giving simple poles at $s=0$ and $s=1$. Equivalently, the standard entire completion is $\frac{1}{2}s(s-1){\pi }^{-s\mathrm{/}2}\mathrm{\Gamma }(s\mathrm{/}2)\zeta (s)\mathrm{.}$ Fact [F3] separately says that $L=\zeta$ itself has only its simple pole at $s=1$.

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.