I finally found the answer, I feel dumb because it is quite simple, but this answer cannot be found at many places on the web.
Consider the gaussian sum of a character modulo $q$ :
$\tau(\chi) = \sum_{n=1}^{q-1} \chi(n) e^{\textstyle\frac{2 i \pi n}{q}}$
I wrote $\sum_{n=1}^{q-1}$ but I could have written $\sum_{n \in G}$ the group of the inversible modulo $q$ (those $n$ with $gcd(n,q)=1$), because $\chi(n) = 0$ if $n \not\in G$. And this is important because it leads to the main trick :
take any $a \in G$ (which is inversible modulo $q$), thus the application $n \to a \,.n$ is a bijection from $G$ to itself, so that : $\forall a \in G, \qquad \tau(\chi) = \sum_{n=1}^{q-1} \chi(a n) e^{\textstyle\frac{2 i \pi a n}{q}}$
and simply because $\chi(a n) = \chi(a) \chi(n)$ : $\tau(\chi) = \chi(a) \sum_{n=1}^{q-1} \chi(n) e^{\textstyle\frac{2 i \pi a n}{q}}$
i.e. :
$\sum_{n=1}^{q-1} \chi(n) e^{\textstyle\frac{2 i \pi a n}{q}} = \tau(\chi) \bar{\chi}(a)$
for $a \in G$, but the uniqueness/inversibility of the Fourier transform implies that the other values of $\chi$'s Fourier transform must be $0$, so that this is true for every $a$.
finally, the discrete Fourier transform (of length $q$) of a Dirichlet character $\chi$ modulo $q$ is $\bar{\chi}\, \tau(\chi)$.
note that in the same way we have also that $\sum_{n=1}^{q-1} \chi(n) e^{-2i \pi n k / q} = \bar{\chi}(k) \overline{\tau(\bar{\chi})}$. then, writing a Fourier series representation for the distribution :
$\delta_\chi(x) = \sum_{n=1}^\infty \chi(n) \delta(x-n) = \frac{1}{q} \sum_{k=1}^\infty \bar{\chi}(k)\left(\tau(\chi) e^{-2 i \pi k x / q} + \overline{\tau(\bar{\chi})} e^{2 i \pi k x / q}\right)$ and with $L(s,\chi) = \int_0^\infty \delta_\chi(x) x^{-s} dx$ we can get the functional equation :
$L(s,\chi) = \sum \chi(n) n^{-s} = L(1-s,\bar{\chi}) A(s)$
with $A(s) = \frac{1}{q}\int_0^\infty \left(\tau(\chi) e^{-2 i \pi x / q} + \overline{\tau(\bar{\chi})} e^{2 i \pi x / q}-2\right) x^{-s}dx$.