Extension of Poisson Summation formula Under the condition f continuous, integrable and:
$|f(t)| + |\hat{f}(t)| \le C (1+|t|)^{-1-a}$ (with a>0)
we have the twisted Poisson formula that holds (where $\chi(n)$ is a primitive Dirichlet character):
$\sum_{-\infty}^{\infty} \chi(n) f(\frac{nx}{\sqrt{q}}) = \frac{A}{x} \sum_{-\infty}^{\infty} \overline{\chi(n)} \widehat{f}(\frac{n}{x \sqrt{q}}) $
In the latest sum the term in zero is null as the character is primitve ($\chi(0)=0$) so I think this equation also holds with some functions integrable but not defined in zero like: $|x|^{-s} e^{-|x|}$ (with $0<s<1$).
Do you have any idea of a reference where I could find such result? (I did not find reference in litterature with such extension of Poisson formula, it seems it does not exists?)
What would be the simplest way to demonstrate it? 
(I thought about multiplying previous function by a function $g_n$ to use classical Poisson summation formula and then make $n \to \infty$, but any idea simpler than that?)
 A: The correct formula is, for $\chi$ primitive of conductor $q$,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)\hat f\left(\frac{n/x}{\sqrt{q}}\right). $$
Here $A:=\sqrt{q}/\tau(\bar\chi)$ is the so-called root number, it is of modulus $1$. 
This formula is essentially equivalent to the functional equation of $L(s,\chi)$, but let me provide a direct proof. We start from the well-known formula
$$ \chi(n) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)e\left(\frac{mn}{q}\right), $$
where $e(x)$ abbreviates $e^{2\pi i x}$. See
Davenport: Multiplicative number theory, Chapter IX, equation (6).
Then we can rewrite the left hand side in the formula as
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = 
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m) \sum_{n\in\mathbb{Z}} 
e\left(\frac{mn}{q}\right) f\left(\frac{nx}{\sqrt{q}}\right).$$
Applying the Poission summation formula for the inner sum on the right hand side,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = 
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)
\sum_{k\in\mathbb{Z}}\int_{-\infty}^\infty e\left(\frac{mt}{q}+kt\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
Denoting $n:=m+qk$ on the right hand side, we obtain
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = 
\frac{1}{\tau(\bar\chi)}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt}{q}\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
By a change of variable,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) = 
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt/x}{\sqrt{q}}\right) f(t)\,dt, $$
where $A$ is as above. This is the stated (corrected) formula.
Remark of 10/14/2013. The formula as proved above holds for integrable continuous functions  satisfying $|f(t)|+|\hat f(t)|\ll (1+|t|)^{-1-\delta}$ for some $\delta>0$. I had an Addendum of 10/12/2013 here with a seemingly more relaxed condition for $q>1$, but this condition turned out to be equivalent to the original one by basic facts on the Fourier transform.
A: I wanted to simply make a comment in GH's answer, but it's getting too long. 
Another very similar way to obtain the formula is the following. (EDIT: This proof is essentially the same as the one given by Noam Elkies in a comment under the question.)
After having appropriately normalized, we reduce the formula to showing that
$$
\sum_n f(n)\chi(n) = \frac{\tau(\chi)}{q} \sum_n \hat{f}\left(\frac{n}{q}\right)\bar{\chi}(n) .
$$
(Remark: I use the definition $\hat{f}(\xi) = \int_{\mathbb{R}} f(t) e(-\xi t)dt$, whereas in GH's proof the exponential is $e(\xi t)$. This explains some discrepancies.) 
Now, note that the arithmetic progression $n\equiv a\pmod q$ is a shifted lattice of $\mathbb{R}$. In fact, 
$$
\sum_{n\equiv a\pmod q}f(n) = \frac{1}{q}\sum_{n} e\left(\frac{an}{q}\right) \hat{f}\left(\frac{n}{q}\right),
$$ 
a corollary of the full Poisson summation formula. Then we have that
$$
\sum_{n}f(n)\chi(n)= \sum_{a=1}^q \chi(a) \sum_{n\equiv a\pmod q} f(n) 
= \frac{1}{q} \sum_{a=1}^q \chi(a) \sum_{n} e\left(\frac{an}{q}\right) \hat{f}\left(\frac{n}{q}\right)
= \frac{\tau(\chi)}{q} \sum_n \hat{f}\left(\frac{n}{q}\right)\bar{\chi}(n) ,
$$ 
since 
$$
\sum_{a=1}^q \chi(a) e\left(\frac{an}{q}\right) = \bar{\chi}(n)\tau(\chi)
$$
for all $n$ if $\chi$ is a primitive character mod $q$. (A fancy way to read this last formula which might be useful conceptually is "a primitive character $\chi$ is an eigenvector of the operator $\bar{F}$, where $F$ is the Fourier transform mod $q$, with eigenvalue $\overline{\tau(\chi)}$.)
A: $$ \sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}g(\log n)=\sum_{\rho}\frac{h( \gamma)}{\zeta '( \rho )}+\sum_{n=1}^{\infty} \frac{1}{\zeta ' (-2n)} \int_{-\infty}^{\infty}dxg(x)e^{-(2n+1/2)x} $$
Also for the Liouville function  we have
$$ \sum_{n=1}^\infty \frac{\lambda(n)}{\sqrt{n}}g(\log n) = \sum_{\rho}\frac{h( \gamma)\zeta(2 \rho )}{\zeta'( \rho)} + \frac{1}{\zeta (1/2)}\int_{-\infty}^\infty dx \, g(x) $$
For the Euler-Phi function the explicit formula reads
$$ \sum_{n=1}^{\infty} \frac{\varphi (n)}{\sqrt{n}}g(\log n)=  \frac{6}{\pi ^2} \int_{-\infty}^\infty dx \, g(x) e^{3x/2} + \sum_\rho \frac{h( \gamma)\zeta(\rho -1 )}{\zeta '( \rho)}+ \frac{1}{2}\sum_{n=1}^\infty \frac{\zeta (-2n-1)}{\zeta'(-2n)} \int_{-\infty}^\infty dx \, g(x)e^{-x(2n+1/2)} $$
