Functional equation for primitive Dirichlet L-functions

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.