[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*}[/step]

[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: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.[/step]

[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: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*}[/step]

[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: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*}[/step]

[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: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 $\varepsilon(\chi)$.[/step]

[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]