Because I was going for a self-driving tour on China National Highway 318 in August, there were no updates since then.

The functional equation for the Riemann zeta function is essentially an application of Jacobi's imaginary transformation of $\vartheta_3(z\vert\tau)$:

\[\vartheta_3(z\vert\tau)=(-i\tau)^{-\frac12}\exp\left(z^2\over i\pi\tau\right)\vartheta_3\left(-\frac z\tau\left\vert-\frac1\tau\right.\right).\tag1\]

In this article, we study a generalization of $\vartheta_3(z\vert\tau)$ to Dirichlet characters that allows us to deduce the functional equation for Dirichlet $L$-functions.

Theta function twisted by a Dirichlet character

Let $\chi$ be a primitive character modulo $k>1$. We define the $\chi$-theta function by

\[\Theta(z\vert\tau,\chi)=\sum_{n\in\mathbb Z}\chi(n)e^{(2\pi inz+\pi in^2\tau)/k}.\tag2\]

Although the summation in (2) is over all integers, we can use the multiplicative property of $\chi$ and the fact that $\chi(0)=0$ to reduce the summation to be over natural numbers. Specifically, we have

\[\Theta(z\vert\tau,\chi)=2\sum_{n\ge1}\chi(n)\cos\left(2\pi nz\over k\right)e^{\pi in^2\tau/k}\tag3\]

when $\chi(-1)=1$ and

\[\Theta(z\vert\tau,\chi)=2i\sum_{n\ge1}\chi(n)\sin\left(2\pi nz\over k\right)e^{\pi in^2\tau/k}\tag4\]

when $\chi(-1)=-1$.

Transformation formula

For the sake of convenience, we define

\[\theta_3(z\vert\tau)=\vartheta_3(\pi z\vert\tau)=\sum_{n\in\mathbb Z}e^{2\pi inz+\pi in^2\tau},\]

so that (1) becomes

\[\theta_3(z\vert\tau)=(-i\tau)^{-\frac12}e^{\pi iz^2\tau'}\theta_3(z\tau'\vert\tau'),\quad\tau'=-\frac1\tau\tag5\]

Let $G(n,\chi)$ denote the character sum

\[G(n,\chi)=\sum_{h\pmod k}\chi(h)e^{2\pi inh/k}.\]

When $\chi$ is primitive, we have $G(n,\chi)=\overline\chi(n)G(1,\chi)$ for all $n$, so that

\[\begin{aligned} G(1,\overline\chi)\Theta(z\vert\tau,\chi) &=\sum_{h\pmod k}\overline\chi(h)\sum_{n\in\mathbb Z}e^{2\pi in(h+z)/k+\pi in^2\tau/k} \\ &=\sum_{h\pmod k}\overline\chi(h)\theta_3\left({(h+z)\over k}\left\vert\frac\tau k\right.\right). \end{aligned}\]

Let $\tau'=-1/\tau$, so $(\tau/k)'=k\tau'$ Then it follows from (5) that

\[\begin{aligned} G(1,\overline\chi)\Theta(z\vert\tau,\chi) &=k^{\frac12}(-i\tau)^{-\frac12}\sum_{h\pmod k}\overline\chi(h)e^{\pi i(h+z)^2\tau'/k}\theta_3((h+z)\tau'\vert k\tau') \\ &=k^{\frac12}(-i\tau)^{-\frac12}\sum_{h\pmod k}\overline\chi(h)\sum_{n\in\mathbb Z}e^{\pi i\tau'[(h+z)^2+2nk(h+z)+n^2k^2]/k} \\ &=k^{\frac12}(-i\tau)^{-\frac12}\sum_{h\pmod k}\overline\chi(h)\sum_{n\in\mathbb Z}e^{\pi i\tau'(nk+h+z)^2/k} \\ &=k^{\frac12}(-i\tau)^{-\frac12}\sum_{m\in\mathbb Z}\overline\chi(m)e^{\pi i\tau'(m+z)^2/k}. \end{aligned}\]

Finally, by expanding $(m+z)^2$ and applying (2), we obtain the transformation formula for $\Theta(z\vert\tau,\chi)$:

\[\Theta(z\vert\tau,\chi)={k^{\frac12}\over G(1,\overline\chi)}(-i\tau)^{-\frac12}e^{\pi i\tau'z^2/k}\Theta(z\tau'\vert\tau',\overline\chi).\tag6\]

Having obtained the transformation formula for $\Theta(z\vert\tau,\chi)$, we now apply it to study the Dirichlet $L$-functions.

Functional equation for Dirichlet $L$-functions

When $\beta>0$ and $\Re(\alpha)>0$, it is known that

\[\int_0^\infty x^{\alpha-1}e^{-\beta x}\mathrm dx=\Gamma(\alpha)\beta^{-\alpha},\]

so we have

\[k^{\frac s2}\pi^{-\frac s2}\Gamma\left(\frac s2\right)L(s,\chi)=\int_0^\infty x^{\frac s2-1}\left(\sum_{n\ge1}\chi(n)e^{-n^2\pi x/k}\right)\mathrm dx.\tag7\]

Comparing (3) and (4), we found that (7) can be connected to $\Theta(z\vert\tau,\chi)$ when $\chi(-1)=1$, so we proceed with this assumption first and discuss the case where $\chi(-1)=-1$ later.

The case where $\chi(-1)=1$

Plugging $z=0$ into (3), we have

\[k^{\frac s2}\pi^{-\frac s2}\Gamma\left(\frac s2\right)L(s,\chi)=\frac12\int_0^\infty x^{\frac s2-1}\Theta(0\vert ix,\chi)\mathrm dx.\tag8\]

Similar to how $\zeta(s)$ was handled, we divide the interval of integration into $[0,1)$ and $[1,+\infty)$ and apply the $z=0$ case of (6) to $[0,1)$:

\[\begin{aligned} \int_0^1 x^{\frac s2-1}\Theta(0\vert ix,\chi)\mathrm dx &={k^{\frac12}\over G(1,\overline\chi)}\int_0^1 x^{ {s-1\over2}-1}\Theta(0\vert ix^{-1},\overline\chi)\mathrm dx \\ &={k^{\frac12}\over G(1,\overline\chi)}\int_1^\infty x^{ {1-s\over2}-1}\Theta(0\vert ix,\overline\chi)\mathrm dx. \end{aligned}\]

Therefore, (8) becomes

\[\begin{aligned} k^{\frac s2}\pi^{-\frac s2}\Gamma\left(\frac s2\right)L(s,\chi) &={k^{\frac12}\over2G(1,\overline\chi)}\int_1^\infty x^{ {1-s\over2}-1}\Theta(0\vert ix,\overline\chi)\mathrm dx \\ &+\frac12\int_1^\infty x^{\frac s2-1}\Theta(0\vert ix,\chi)\mathrm dx. \end{aligned}\tag9\]

Notice that

\[G(1,\overline\chi)=\chi(-1)\overline{G(1,\chi)}.\tag{10}\]

and $\vert G(1,\chi)\vert=k^{\frac12}$, so when $\chi(-1)=1$, there is

\[\begin{aligned} {k^{\frac12}\over G(1,\chi)}k^{\frac s2}\pi^{-\frac s2}\Gamma\left(\frac s2\right)L(s,\chi) &=\frac12\int_1^\infty x^{ {1-s\over2}-1}\Theta(0\vert ix,\overline\chi)\mathrm dx \\ &+{k^{\frac12}\over2 G(1,\chi)}\int_1^\infty x^{\frac s2-1}\Theta(0\vert ix,\chi)\mathrm dx. \end{aligned}\]

Comparing the right-hand side with (9), we obtain the functional equation for $L(s,\chi)$ when $\chi(-1)=1$:

\[k^{1-s\over2}\pi^{-{1-s\over2}}\Gamma\left(1-s\over2\right)L(1-s,\overline\chi)={k^{\frac12}\over G(1,\chi)}k^{\frac s2}\pi^{-\frac s2}\Gamma\left(\frac s2\right)L(s,\chi).\tag{11}\]

Having completed the case for $\chi(-1)=1$, we now look at how the $\chi(-1)=-1$ case is handled.

The case where $\chi(-1)=-1$

Because (4) is used instead of (3) when $\chi(-1)=-1$, we find that $\Theta(0\vert\tau,\chi)=0$, so we cannot proceed from (7) like how we did in the previous section.

Since the vanishing is due to the fact that $\Theta(z\vert\tau,\chi)$ is an odd function of $z$ when $\chi(-1)=-1$, so its $z$-derivative $\Theta'(z\vert\tau,\chi)$ will not vanish at $z=0$:

\[\Theta'(z\vert\tau,\chi)={4\pi i\over k}\sum_{n\ge1}\chi(n)\cos\left(2\pi nz\over k\right)ne^{\pi in^2\tau/k}.\]

Differentiating (6) at $z=0$, we have

\[\Theta'(0\vert\tau,\chi)={ik^{\frac12}\over G(1,\overline\chi)}(-i\tau)^{-\frac32}\Theta'(0\vert\tau,\chi).\tag{12}\]

Similar to how (11) is proven via the $z=0$ case of (6), we use (12) to deduce the following:

\[\begin{aligned} k^{s+1\over2}\pi^{-{s+1\over2}} &\Gamma\left(s+1\over2\right)L(s,\chi) ={k\over4\pi i}\int_0^\infty x^{ {s+1\over2}-1}\Theta'(0\vert ix,\chi)\mathrm dx \\ &={ik^{\frac12}\over G(1,\overline\chi)}{k\over4\pi i}\int_1^\infty x^{ {(1-s)+1\over2}-1}\Theta'(0\vert ix,\chi)\mathrm dx \\ &+{k\over4\pi i}\int_1^\infty x^{ {s+1\over2}-1}\Theta'(0\vert ix,\chi)\mathrm dx. \end{aligned}\]

Finally, observe that (10) becomes $G(1,\overline\chi)=-\overline{G(1,\chi)}$ when $\chi(-1)=-1$, so we obtain the functional equation for $L(s,\chi)$ when $\chi(-1)=-1$:

\[k^{2-s\over2}\pi^{-{2-s\over2}}\Gamma\left(2-s\over2\right)L(1-s,\overline\chi)={ik^{\frac12}\over G(1,\chi)}k^{s+1\over2}\pi^{-{s+1\over2}}\Gamma\left(s+1\over2\right)L(s,\chi).\tag{13}\]

Unification

Remembering two functional equations according to the values of $\chi(-1)$ would be inconvenient when we wish to study the properties of $L(s,\chi)$ in general, so we define a variable

\[a={1-\chi(-1)\over2}= \begin{cases} 0 & \chi(-1)=1, \\ 1 & \chi(-1)=-1 \end{cases}\]

and a function

\[\xi(s,\chi)=k^{ {s+a\over2}}\pi^{-{s+a\over2}}\Gamma\left(s+a\over2\right)L(s,\chi),\]

so (11) and (13) can be written as

\[\xi(1-s,\overline\chi)={i^ak^{\frac12}\over G(1,\chi)}\xi(s,\chi).\tag{14}\]

This is consistent with (14) in §9 of Davenport's Multiplicative Number Theory.

Alternative form

Although (14) is a symmetric form of the functional equation, due to the presence of Gamma factors on both sides, it is inconvenient for us to connect the asymptotic behavior of $L(s,\chi)$ and $L(1-s,\overline\chi)$. Thus, to achieve this purpose, we isolate $L(1-s,\overline\chi)$ in (14):

\[\begin{aligned} L(1-s,\overline\chi) &={i^ak^{\frac12}\over G(1,\chi)}k^{2s-1\over2}\pi^{-{2s-1\over2}}{\Gamma\left(s+a\over2\right)\over\Gamma\left(1-s+a\over2\right)}L(s,\chi) \\ &={i^a\over G(1,\chi)}k^s\pi^{\frac12-s}{\Gamma\left(s+a\over2\right)\over\Gamma\left(1-{1-a+s\over2}\right)}L(s,\chi). \end{aligned}\tag{15}\]

To simplify the Gamma functions, we use Euler's reflection formula:

\[\Gamma(z)\Gamma(1-z)={\pi\over\sin(\pi z)}\]

and Legendre's duplication formula:

\[\Gamma(z)\Gamma\left(z+\frac12\right)=2^{1-2z}\pi^{\frac12}\Gamma(2z),\]

so we have

\[\begin{aligned} {\Gamma\left(s+a\over2\right)\over\Gamma\left(1-{1-a+s\over2}\right)} &={\Gamma\left(s+a\over2\right)\Gamma\left(1-a+s\over2\right)\over\pi}\sin \left(\pi(1-a+s)\over2\right) \\ &=\pi^{-1}{\Gamma\left(\frac s2\right)\Gamma\left(\frac s2+\frac12\right)}\sin \left(\pi(1-a+s)\over2\right) \\ &=\pi^{-\frac12}2^{-s}\Gamma(s)2\sin \left(\pi(1-a+s)\over2\right). \end{aligned}\]

Furthermore, because

\[\begin{aligned} 2\sin \left(\pi(1-a+s)\over2\right) &={i^{1-a}e^{\pi is/2}-i^{a-1}e^{-\pi is}\over i} \\ &=i^{-a}e^{\pi is/2}+i^ae^{-\pi is/2}, \end{aligned}\]

we can transform (15) into

\[L(1-s,\overline\chi)={k^s\Gamma(s)\over(2\pi)^s}(e^{\pi is/2}+i^{2a}e^{-\pi is/2}){L(s,\chi)\over G(1,\chi)}.\]

Now, combining (10) and the fact that $i^{2a}=\chi(-1)$, we obtain

\[L(1-s,\overline\chi)={k^{s-1}\Gamma(s)\over(2\pi)^s}(e^{\pi is/2}+\chi(-1)e^{-\pi is/2})\chi(-1)G(1,\overline\chi)L(s,\chi).\]

Simplifying and swapping the positions of $\chi$ and $\overline\chi$, we obtain a convenient form of the functional equation

\[L(1-s,\chi)={k^{s-1}\Gamma(s)\over(2\pi)^s}(e^{-\pi is/2}+\chi(-1)e^{\pi is/2})G(1,\chi)L(s,\overline\chi),\tag{16}\]

which is consistent with Theorem 12.11 of Apostol's Introduction to Analytic Number Theory.

Notes

The transformation identities (6) and (12) can be proven directly using Poisson summation formula. This is often the approach adopted in textbooks (e.g. Apostol, Davenport, and Montgomery & Vaughan) as it requires no use of theta functions, but introducing $\Theta(z\vert\tau,\chi)$ reveals more motivations behind the derivation of the functional equations.

Although we deduce it from the symmetric form (14), the standard proof of (16) uses a transformation formula of the Hurwitz zeta function defined by

\[\zeta(s,a)=\sum_{n\ge0}{1\over(n+a)^s}.\]

Details of this approach are found in §12 of Apostol.