[proofplan]
We prove the functional equation by applying Poisson summation to a theta series twisted by the primitive character $\chi$. The finite [Fourier transform](/page/Fourier%20Transform) that appears in Poisson summation is evaluated by the primitive Gauss sum identity, giving the theta transformation law with the constant $i^{-a}\tau(\chi)/\sqrt q$. Taking the Mellin transform of the transformed theta series produces the completed $L$-function, and splitting the Mellin integral at $1$ converts the transformation $t \mapsto 1/t$ into the symmetry $s \mapsto 1-s$. The principal character case contributes the usual constant terms and therefore gives meromorphic, rather than entire, continuation.
[/proofplan]
[step:Define the twisted theta series and record its Poisson transformation]
Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Define the twisted theta series as the map
\begin{align*}
\Theta_a(\cdot,\chi):(0,\infty)&\to \mathbb{C} \\
t&\mapsto \sum_{n \in \mathbb{Z}} n^a \chi(n)\exp\left(-\frac{\pi n^2 t}{q}\right),
\end{align*}
where $0^0\chi(0)$ is interpreted as $\chi(0)$ when $a=0$, and $0^1\chi(0)$ is interpreted as $0$ when $a=1$.
We use the [Poisson summation formula](/theorems/902) for Schwartz functions on $\mathbb{R}$ and the [Fourier Transform of a Primitive Dirichlet Character](/theorems/TEMP-51), which gives the primitive Gauss sum identity
\begin{align*}
\sum_{r=1}^{q}\chi(r)e^{2\pi i mr/q}
=
\overline{\chi}(m)\tau(\chi)
\qquad (m \in \mathbb{Z}).
\end{align*}
For $t>0$, define the Schwartz function
\begin{align*}
f_t:\mathbb{R} &\to \mathbb{C} \\
x &\mapsto x^a\exp\left(-\frac{\pi x^2t}{q}\right).
\end{align*}
Decomposing integers into residue classes modulo $q$ gives
\begin{align*}
\Theta_a(t,\chi)
=
\sum_{r=1}^{q}\chi(r)\sum_{k \in \mathbb{Z}}(r+kq)^a\exp\left(-\frac{\pi (r+kq)^2t}{q}\right).
\end{align*}
For each $r \in \{1,\dots,q\}$, applying Poisson summation to the map $x \mapsto f_t(r+qx)$ gives
\begin{align*}
\sum_{k \in \mathbb{Z}}f_t(r+kq)
=
\frac{1}{q}\sum_{m \in \mathbb{Z}}e^{2\pi i mr/q}\widehat{f_t}\left(\frac{m}{q}\right),
\end{align*}
where the Fourier transform is the map
\begin{align*}
\widehat{f_t}:\mathbb R&\to\mathbb C\\
\xi&\mapsto
\int_{\mathbb{R}}f_t(x)e^{-2\pi i x\xi}\,d\mathcal{L}^1(x).
\end{align*}
The Gaussian Fourier transform and differentiation identity give, for $m \in \mathbb{Z}$,
\begin{align*}
\widehat{f_t}\left(\frac{m}{q}\right)
&=
\begin{cases}
q^{1/2}t^{-1/2}\exp\left(-\dfrac{\pi m^2}{qt}\right), & a=0, \\
-iq^{1/2}t^{-3/2}m\exp\left(-\dfrac{\pi m^2}{qt}\right), & a=1,
\end{cases} \\
&=
i^{-a}q^{1/2}t^{-a-1/2}m^a\exp\left(-\frac{\pi m^2}{qt}\right).
\end{align*}
Substituting this into the residue-class decomposition and then using the primitive Gauss sum identity yields
\begin{align*}
\Theta_a(t,\chi)
&=
\frac{i^{-a}}{\sqrt q}t^{-a-1/2}
\sum_{m \in \mathbb{Z}}m^a\exp\left(-\frac{\pi m^2}{qt}\right)
\sum_{r=1}^{q}\chi(r)e^{2\pi i mr/q} \\
&=
\frac{i^{-a}\tau(\chi)}{\sqrt q}\,
t^{-a-1/2}
\sum_{m \in \mathbb{Z}}m^a\overline{\chi}(m)
\exp\left(-\frac{\pi m^2}{qt}\right) \\
&=
\frac{i^{-a}\tau(\chi)}{\sqrt q}\,
t^{-a-1/2}\Theta_a(t^{-1},\overline{\chi})
\qquad (t>0).
\end{align*}
Thus, with
\begin{align*}
\varepsilon(\chi):=\frac{i^{-a}\tau(\chi)}{\sqrt q},
\end{align*}
we have
\begin{align*}
\Theta_a(t,\chi)
=
\varepsilon(\chi)t^{-a-1/2}\Theta_a(t^{-1},\overline{\chi}).
\end{align*}
[guided]
The point of introducing $\Theta_a(t,\chi)$ is that it packages the parity of $\chi$ into a theta series whose Mellin transform is exactly the completed $L$-function. The factor $n^a$ is present only in the odd case; it makes the summand even in $n$, because
\begin{align*}
(-n)^a\chi(-n)
=
(-1)^a n^a \chi(-1)\chi(n)
=
(-1)^a(-1)^a n^a\chi(n)
=
n^a\chi(n).
\end{align*}
For $t>0$, define
\begin{align*}
f_t:\mathbb{R} &\to \mathbb{C} \\
x &\mapsto x^a\exp\left(-\frac{\pi x^2t}{q}\right).
\end{align*}
This is a Schwartz function, so Poisson summation applies after we split the integer sum into residue classes modulo $q$. Namely,
\begin{align*}
\Theta_a(t,\chi)
=
\sum_{r=1}^{q}\chi(r)\sum_{k \in \mathbb{Z}}f_t(r+kq).
\end{align*}
For fixed $r \in \{1,\dots,q\}$, Poisson summation applied to the Schwartz map $x \mapsto f_t(r+qx)$ gives
\begin{align*}
\sum_{k \in \mathbb{Z}}f_t(r+kq)
=
\frac{1}{q}\sum_{m \in \mathbb{Z}}e^{2\pi i mr/q}\widehat{f_t}\left(\frac{m}{q}\right),
\end{align*}
where the Fourier transform is the map
\begin{align*}
\widehat{f_t}:\mathbb R&\to\mathbb C\\
\xi&\mapsto
\int_{\mathbb{R}}f_t(x)e^{-2\pi i x\xi}\,d\mathcal{L}^1(x).
\end{align*}
We now compute this Fourier transform, since it is the source of the exact power of $t$ and the phase $i^{-a}$. If $a=0$, the Gaussian transform gives
\begin{align*}
\widehat{f_t}(\xi)
=
\left(\frac{q}{t}\right)^{1/2}\exp\left(-\frac{\pi q\xi^2}{t}\right),
\end{align*}
and hence
\begin{align*}
\widehat{f_t}\left(\frac{m}{q}\right)
=
q^{1/2}t^{-1/2}\exp\left(-\frac{\pi m^2}{qt}\right).
\end{align*}
If $a=1$, we use
\begin{align*}
\frac{d}{d\xi}\widehat{g}(\xi)
=
\int_{\mathbb{R}}g(x)(-2\pi ix)e^{-2\pi ix\xi}\,d\mathcal{L}^1(x)
\end{align*}
for the Gaussian map $g:\mathbb{R}\to\mathbb{C}$, $x\mapsto \exp(-\pi x^2t/q)$. Therefore
\begin{align*}
\widehat{f_t}(\xi)
&=
-\frac{1}{2\pi i}\frac{d}{d\xi}\left[\left(\frac{q}{t}\right)^{1/2}\exp\left(-\frac{\pi q\xi^2}{t}\right)\right] \\
&=
-iq^{3/2}t^{-3/2}\xi\exp\left(-\frac{\pi q\xi^2}{t}\right),
\end{align*}
so
\begin{align*}
\widehat{f_t}\left(\frac{m}{q}\right)
=
-iq^{1/2}t^{-3/2}m\exp\left(-\frac{\pi m^2}{qt}\right).
\end{align*}
Both cases are summarized as
\begin{align*}
\widehat{f_t}\left(\frac{m}{q}\right)
=
i^{-a}q^{1/2}t^{-a-1/2}m^a\exp\left(-\frac{\pi m^2}{qt}\right).
\end{align*}
Substitution into Poisson summation gives
\begin{align*}
\Theta_a(t,\chi)
=
\frac{i^{-a}}{\sqrt q}t^{-a-1/2}
\sum_{m \in \mathbb{Z}}m^a\exp\left(-\frac{\pi m^2}{qt}\right)
\sum_{r=1}^{q}\chi(r)e^{2\pi i mr/q}.
\end{align*}
The finite Fourier transform of the primitive character is computed by the [Fourier Transform of a Primitive Dirichlet Character](/theorems/TEMP-51):
\begin{align*}
\sum_{r=1}^{q}\chi(r)e^{2\pi i mr/q}
=
\overline{\chi}(m)\tau(\chi).
\end{align*}
Thus
\begin{align*}
\Theta_a(t,\chi)
=
\frac{i^{-a}\tau(\chi)}{\sqrt q}\,
t^{-a-1/2}\Theta_a(t^{-1},\overline{\chi})
\qquad (t>0).
\end{align*}
Thus the theta series transforms under $t \mapsto t^{-1}$ with multiplier
\begin{align*}
\varepsilon(\chi):=\frac{i^{-a}\tau(\chi)}{\sqrt q}.
\end{align*}
[/guided]
[/step]
[step:Compute the Mellin transform of the twisted theta series]
Assume first that $\chi$ is non-principal. Then $\chi(0)=0$. For $\operatorname{Re}(s)>1$, the series
\begin{align*}
\sum_{n=1}^{\infty}|\chi(n)|n^a
\int_0^\infty
\exp\left(-\frac{\pi n^2t}{q}\right)t^{\operatorname{Re}(s+a)/2-1}\,d\mathcal{L}^1(t)
\end{align*}
converges, because its value is a constant multiple of $\sum_{n=1}^{\infty}|\chi(n)|n^{-\operatorname{Re}(s)}$. Hence Tonelli's theorem applied to the absolute-value majorant permits termwise integration:
\begin{align*}
\int_0^\infty
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
&=
2\sum_{n=1}^{\infty} n^a\chi(n)
\int_0^\infty
\exp\left(-\frac{\pi n^2t}{q}\right)t^{(s+a)/2-1}\,d\mathcal{L}^1(t) \\
&=
2\sum_{n=1}^{\infty} n^a\chi(n)
\left(\frac{q}{\pi n^2}\right)^{(s+a)/2}
\Gamma\left(\frac{s+a}{2}\right) \\
&=
2\left(\frac{q}{\pi}\right)^{(s+a)/2}
\Gamma\left(\frac{s+a}{2}\right)
\sum_{n=1}^{\infty}\frac{\chi(n)}{n^s} \\
&=
2\Lambda(s,\chi).
\end{align*}
The second equality uses the change of variables
\begin{align*}
u=\frac{\pi n^2t}{q},
\qquad
d\mathcal{L}^1(u)=\frac{\pi n^2}{q}\,d\mathcal{L}^1(t),
\end{align*}
in the defining integral for the gamma function.
[guided]
We now connect the theta series to $L(s,\chi)$. Since $\chi$ is non-principal in this step, $\chi(0)=0$, so there is no constant term in the theta series. Also, the summand is even in $n$, as checked in the previous step. Hence
\begin{align*}
\Theta_a(t,\chi)
=
2\sum_{n=1}^{\infty} n^a\chi(n)\exp\left(-\frac{\pi n^2t}{q}\right).
\end{align*}
For $\operatorname{Re}(s)>1$, the series and integral are absolutely convergent, so termwise integration is justified by Tonelli's theorem applied to the absolutely convergent majorant.
For each $n \in \mathbb{N}$, evaluate the integral
\begin{align*}
\int_0^\infty
\exp\left(-\frac{\pi n^2t}{q}\right)t^{(s+a)/2-1}\,d\mathcal{L}^1(t).
\end{align*}
Use the substitution
\begin{align*}
u=\frac{\pi n^2t}{q},
\qquad
t=\frac{qu}{\pi n^2},
\qquad
d\mathcal{L}^1(t)=\frac{q}{\pi n^2}\,d\mathcal{L}^1(u).
\end{align*}
Then
\begin{align*}
\int_0^\infty
\exp\left(-\frac{\pi n^2t}{q}\right)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
&=
\left(\frac{q}{\pi n^2}\right)^{(s+a)/2}
\int_0^\infty e^{-u}u^{(s+a)/2-1}\,d\mathcal{L}^1(u) \\
&=
\left(\frac{q}{\pi n^2}\right)^{(s+a)/2}
\Gamma\left(\frac{s+a}{2}\right).
\end{align*}
Substituting this into the Mellin integral gives
\begin{align*}
\int_0^\infty
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
=
2\left(\frac{q}{\pi}\right)^{(s+a)/2}
\Gamma\left(\frac{s+a}{2}\right)
\sum_{n=1}^{\infty}\frac{\chi(n)}{n^s}.
\end{align*}
The final sum is $L(s,\chi)$ by definition, so the Mellin transform is $2\Lambda(s,\chi)$.
[/guided]
[/step]
[step:Split the Mellin integral and change variables to obtain the functional equation]
For the non-principal case define the map
\begin{align*}
A_\chi:\mathbb{C}&\to \mathbb{C} \\
s&\mapsto
\int_1^\infty
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t).
\end{align*}
The exponential decay of $\Theta_a(t,\chi)$ as $t \to \infty$ implies that $A_\chi(s)$ is an entire function of $s$.
Using the theta transformation from the first step on the interval $(0,1)$ gives
\begin{align*}
\int_0^1
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
&=
\varepsilon(\chi)
\int_0^1
\Theta_a(t^{-1},\overline{\chi})t^{(s-a-3)/2}\,d\mathcal{L}^1(t).
\end{align*}
Now substitute $u=t^{-1}$. Then $t=u^{-1}$ and $d\mathcal{L}^1(t)=u^{-2}\,d\mathcal{L}^1(u)$ after reversing the limits. Therefore
\begin{align*}
\int_0^1
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
&=
\varepsilon(\chi)
\int_1^\infty
\Theta_a(u,\overline{\chi})
u^{(1-s+a)/2-1}\,d\mathcal{L}^1(u) \\
&=
\varepsilon(\chi)A_{\overline{\chi}}(1-s).
\end{align*}
Combining this with the Mellin identity gives
\begin{align*}
2\Lambda(s,\chi)
=
A_\chi(s)+\varepsilon(\chi)A_{\overline{\chi}}(1-s).
\end{align*}
The same formula applied to $\overline{\chi}$ gives
\begin{align*}
2\Lambda(1-s,\overline{\chi})
=
A_{\overline{\chi}}(1-s)+\varepsilon(\overline{\chi})A_\chi(s).
\end{align*}
The [Magnitude of a Primitive Dirichlet Gauss Sum](/theorems/TEMP-52) gives $|\tau(\chi)|^2=q$. Also,
\begin{align*}
\tau(\overline{\chi})
&=
\sum_{r=1}^{q}\overline{\chi}(r)e^{2\pi i r/q} \\
&=
\chi(-1)\sum_{u=1}^{q}\overline{\chi}(u)e^{-2\pi i u/q} \\
&=
\chi(-1)\overline{\tau(\chi)},
\end{align*}
where the second equality uses the substitution $u\equiv -r\pmod q$. Therefore
\begin{align*}
\tau(\chi)\tau(\overline{\chi})=\chi(-1)q=(-1)^a q,
\end{align*}
and hence
\begin{align*}
\varepsilon(\chi)\varepsilon(\overline{\chi})
=
\frac{i^{-a}i^{-a}\tau(\chi)\tau(\overline{\chi})}{q}
=
i^{-2a}(-1)^a
=
1.
\end{align*}
Multiplying the second displayed formula by $\varepsilon(\chi)$ and using $\varepsilon(\chi)\varepsilon(\overline{\chi})=1$ gives
\begin{align*}
2\varepsilon(\chi)\Lambda(1-s,\overline{\chi})
=
\varepsilon(\chi)A_{\overline{\chi}}(1-s)+A_\chi(s)
=
2\Lambda(s,\chi).
\end{align*}
Thus
\begin{align*}
\Lambda(s,\chi)
=
\varepsilon(\chi)\Lambda(1-s,\overline{\chi}).
\end{align*}
[guided]
The Mellin transform identity was first proved for $\operatorname{Re}(s)>1$. To continue it, split the integral into the part from $1$ to $\infty$ and the part from $0$ to $1$. Define the map
\begin{align*}
A_\chi:\mathbb{C}&\to \mathbb{C} \\
s&\mapsto
\int_1^\infty
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t).
\end{align*}
Because the Gaussian factor $\exp(-\pi n^2t/q)$ decays exponentially as $t \to \infty$, this integral converges locally uniformly for all $s \in \mathbb{C}$. Therefore $A_\chi$ is entire.
The remaining part is the integral over $(0,1)$. Here the theta transformation is designed exactly to replace small $t$ by large $t$:
\begin{align*}
\Theta_a(t,\chi)
=
\varepsilon(\chi)t^{-a-1/2}\Theta_a(t^{-1},\overline{\chi}).
\end{align*}
Substituting this identity gives
\begin{align*}
\int_0^1
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
=
\varepsilon(\chi)
\int_0^1
\Theta_a(t^{-1},\overline{\chi})t^{(s-a-3)/2}\,d\mathcal{L}^1(t).
\end{align*}
Now set $u=t^{-1}$. Then the domain $(0,1)$ becomes $(\infty,1)$, and reversing the limits produces the integral over $(1,\infty)$. Since
\begin{align*}
d\mathcal{L}^1(t)=u^{-2}\,d\mathcal{L}^1(u),
\end{align*}
we obtain
\begin{align*}
t^{(s-a-3)/2}\,d\mathcal{L}^1(t)
=
u^{-(s-a-3)/2}u^{-2}\,d\mathcal{L}^1(u)
=
u^{(1-s+a)/2-1}\,d\mathcal{L}^1(u).
\end{align*}
Therefore
\begin{align*}
\int_0^1
\Theta_a(t,\chi)t^{(s+a)/2-1}\,d\mathcal{L}^1(t)
=
\varepsilon(\chi)A_{\overline{\chi}}(1-s).
\end{align*}
Combining this with the Mellin transform calculation yields
\begin{align*}
2\Lambda(s,\chi)
=
A_\chi(s)+\varepsilon(\chi)A_{\overline{\chi}}(1-s).
\end{align*}
The same reasoning applied to the conjugate character $\overline{\chi}$ gives
\begin{align*}
2\Lambda(1-s,\overline{\chi})
=
A_{\overline{\chi}}(1-s)+\varepsilon(\overline{\chi})A_\chi(s).
\end{align*}
To compare these two formulas, we need the product of the two root numbers. The [Magnitude of a Primitive Dirichlet Gauss Sum](/theorems/TEMP-52) gives
\begin{align*}
|\tau(\chi)|^2=q.
\end{align*}
The conjugate Gauss sum is computed directly:
\begin{align*}
\tau(\overline{\chi})
&=
\sum_{r=1}^{q}\overline{\chi}(r)e^{2\pi i r/q} \\
&=
\chi(-1)\sum_{u=1}^{q}\overline{\chi}(u)e^{-2\pi i u/q} \\
&=
\chi(-1)\overline{\tau(\chi)}.
\end{align*}
Therefore
\begin{align*}
\tau(\chi)\tau(\overline{\chi})=\chi(-1)q=(-1)^a q.
\end{align*}
Hence
\begin{align*}
\varepsilon(\chi)\varepsilon(\overline{\chi})
=
\frac{i^{-a}i^{-a}\tau(\chi)\tau(\overline{\chi})}{q}
=
i^{-2a}(-1)^a
=
1.
\end{align*}
Multiplying the formula for $\Lambda(1-s,\overline{\chi})$ by $\varepsilon(\chi)$ now gives exactly the formula for $\Lambda(s,\chi)$:
\begin{align*}
2\varepsilon(\chi)\Lambda(1-s,\overline{\chi})
=
\varepsilon(\chi)A_{\overline{\chi}}(1-s)+A_\chi(s)
=
2\Lambda(s,\chi).
\end{align*}
Dividing by $2$ proves the functional equation.
[/guided]
[/step]
[step:Handle the principal primitive character and identify the poles]
If $\chi$ is principal and primitive, then
\begin{align*}
q&=1,\\
a&=0,\\
L(s,\chi)&=\zeta(s),
\end{align*}
where $\zeta$ denotes the Riemann zeta function
\begin{align*}
\zeta:\{s \in \mathbb{C}:\operatorname{Re}(s)>1\}&\to \mathbb{C} \\
s&\mapsto \sum_{n=1}^{\infty}n^{-s}.
\end{align*}
The theta series is
\begin{align*}
\Theta_0(t,\chi)
=
\sum_{n \in \mathbb{Z}}e^{-\pi n^2t},
\end{align*}
and its constant term is $1$. The Mellin transform identity becomes
\begin{align*}
2\Lambda(s,\chi)
=
\int_0^\infty
\bigl(\Theta_0(t,\chi)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t).
\end{align*}
Splitting at $1$, using
\begin{align*}
\Theta_0(t,\chi)=t^{-1/2}\Theta_0(t^{-1},\chi),
\end{align*}
and changing variables $u=t^{-1}$ gives
\begin{align*}
2\Lambda(s,\chi)
=
B(s)+B(1-s)+\frac{2}{s-1}-\frac{2}{s},
\end{align*}
where $B$ is the map
\begin{align*}
B:\mathbb{C}&\to \mathbb{C} \\
s&\mapsto
\int_1^\infty
\bigl(\Theta_0(t,\chi)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t).
\end{align*}
The function $B$ is entire by exponential decay. Hence $\Lambda(s,\chi)$ is meromorphic on $\mathbb{C}$, with possible simple poles at $s=0$ and $s=1$, and the displayed expression is invariant under $s \mapsto 1-s$. Since $\tau(\chi)=1$ and $\varepsilon(\chi)=1$ in this case, the functional equation is
\begin{align*}
\Lambda(s,\chi)=\Lambda(1-s,\chi).
\end{align*}
[guided]
The only primitive principal character is the character modulo $1$. Thus
\begin{align*}
q&=1,\\
a&=0,\\
L(s,\chi)&=\zeta(s).
\end{align*}
The theta series is
\begin{align*}
\Theta_0(t,\chi)
=
\sum_{n\in\mathbb Z}e^{-\pi n^2t}.
\end{align*}
Its constant term is $1$, so the Mellin-transform calculation is applied to $\Theta_0(t,\chi)-1$. Splitting the resulting integral at $1$, using
\begin{align*}
\Theta_0(t,\chi)=t^{-1/2}\Theta_0(t^{-1},\chi),
\end{align*}
and changing variables gives
\begin{align*}
2\Lambda(s,\chi)
=
B(s)+B(1-s)+\frac{2}{s-1}-\frac{2}{s},
\end{align*}
where
\begin{align*}
B:\mathbb C&\to\mathbb C\\
s&\mapsto
\int_1^\infty
\bigl(\Theta_0(t,\chi)-1\bigr)t^{s/2-1}\,d\mathcal{L}^1(t).
\end{align*}
The function $B$ is entire by exponential decay. Thus the continuation is meromorphic with possible simple poles at
\begin{align*}
s=0
\qquad\text{and}\qquad
s=1.
\end{align*}
Since
\begin{align*}
\tau(\chi)=1,
\qquad
\varepsilon(\chi)=1,
\end{align*}
the symmetry of the displayed expression gives
\begin{align*}
\Lambda(s,\chi)=\Lambda(1-s,\chi).
\end{align*}
[/guided]
[/step]
[step:Deduce entire continuation in the non-principal case and compute the modulus of the root number]
In the non-principal case, the formula
\begin{align*}
2\Lambda(s,\chi)
=
A_\chi(s)+\varepsilon(\chi)A_{\overline{\chi}}(1-s)
\end{align*}
expresses $\Lambda(s,\chi)$ as a sum of entire functions of $s$. Therefore $\Lambda(s,\chi)$ extends to an entire function on $\mathbb{C}$.
Finally, the primitive Gauss sum magnitude formula gives
\begin{align*}
|\tau(\chi)|=\sqrt q.
\end{align*}
Since $|i^{-a}|=1$, we obtain
\begin{align*}
|\varepsilon(\chi)|
=
\left|\frac{i^{-a}\tau(\chi)}{\sqrt q}\right|
=
\frac{|\tau(\chi)|}{\sqrt q}
=
1.
\end{align*}
This proves the meromorphic continuation, the asserted entireness in the non-principal case, the functional equation, and the modulus statement for $\varepsilon(\chi)$.
[guided]
In the nonprincipal case, the split-Mellin formula is
\begin{align*}
2\Lambda(s,\chi)
=
A_\chi(s)+\varepsilon(\chi)A_{\overline{\chi}}(1-s).
\end{align*}
Both functions
\begin{align*}
A_\chi:\mathbb C&\to\mathbb C,\\
A_{\overline{\chi}}:\mathbb C&\to\mathbb C
\end{align*}
are entire by local [uniform convergence](/page/Uniform%20Convergence) of the exponentially decaying integrals on $[1,\infty)$. Hence $\Lambda(s,\chi)$ is entire in the nonprincipal case.
Finally, by the [Magnitude of a Primitive Dirichlet Gauss Sum](/theorems/TEMP-52),
\begin{align*}
|\tau(\chi)|=\sqrt q.
\end{align*}
Since
\begin{align*}
|i^{-a}|=1,
\end{align*}
we get
\begin{align*}
|\varepsilon(\chi)|
&=
\left|\frac{i^{-a}\tau(\chi)}{\sqrt q}\right|\\
&=
\frac{|\tau(\chi)|}{\sqrt q}\\
&=1.
\end{align*}
This proves the entireness statement in the nonprincipal case and the asserted modulus of the root number.
[/guided]
[/step]
