Step 1: Theta function. Define the theta series $\theta(x, \chi) = \sum_{n=-\infty}^{\infty} \chi(n) n^a e^{-\pi n^2 x/q}$ for $x > 0$. The Poisson summation formula, applied with the Gauss sum $\tau(\chi)$, gives: $\theta(x, \chi) = \frac{\tau(\chi)}{i^a \sqrt{qx}} x^{-(1+a)/2+1/2} \cdot q^{1/2} \theta(1/x, \overline{\chi})$ More precisely: $\theta(x, \chi) = \frac{\tau(\chi)}{i^a q^{1/2}} x^{-1/2} \theta(1/x, \overline{\chi})$.

Step 2: Mellin transform. Compute the Mellin transform: $\int_0^\infty \theta(x, \chi) x^{s/2} \frac{dx}{x} = 2 \left(\frac{q}{\pi}\right)^{(s+a)/2} \Gamma\!\left(\frac{s+a}{2}\right) L(s, \chi) = 2\xi(s, \chi)$ (extracting the $n > 0$ terms and using $\int_0^\infty x^{s/2} e^{-\pi n^2 x/q} dx/x = (\pi n^2/q)^{-s/2}\Gamma(s/2)$).

Step 3: Splitting the integral. Write $\int_0^\infty = \int_0^1 + \int_1^\infty$. In $\int_0^1$, substitute $x \to 1/x$ and use the transformation law from Step 1: $\int_0^1 \theta(x, \chi) x^{s/2} \frac{dx}{x} = \frac{\tau(\chi)}{i^a\sqrt{q}} \int_1^\infty \theta(x, \overline{\chi}) x^{(1-s)/2} \frac{dx}{x}$

Step 4: Combining. Therefore: $2\xi(s, \chi) = \int_1^\infty \theta(x, \chi) x^{s/2} \frac{dx}{x} + \frac{\tau(\chi)}{i^a\sqrt{q}} \int_1^\infty \theta(x, \overline{\chi}) x^{(1-s)/2} \frac{dx}{x}$

The right side is an entire function of $s$ (since $\theta$ decays exponentially for $x \geq 1$). Replacing $s$ by $1-s$ and $\chi$ by $\overline{\chi}$, and noting $\tau(\overline{\chi}) = \overline{\chi(-1)}\overline{\tau(\chi)} = (-1)^a \overline{\tau(\chi)}$: $\xi(s, \chi) = \varepsilon(\chi)\xi(1-s, \overline{\chi})$. $\square$