The discrete Fourier transform of a Dirichlet charachter I usually work in number theory so I am not familiar with Fourier transforms, I have read up on them and know the basics but it never seems to be in number theory language.
I am trying to find the transform of a primitive Dirichlet character $\chi(n) \bmod q$. I know this is a periodic function and $\chi(n)=\exp\left(\frac{Kv(n)}{\phi(p^\alpha)}\right)$ but I have no idea have to find its transform or the transform of $f(n)\chi(n)$
Yes you are right, say how would you calculate $\sum_(n\epsilon Z) f(n)\chi(n)$
 A: 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 $A(s) = \frac{1}{q}\int_0^\infty \left(\tau(\chi) e^{-2 i \pi  x / q} + \overline{\tau(\bar{\chi})} e^{2 i \pi  x / q}-2\right) x^{-s}dx$.
