Let $K$ be an imaginary quadratic field and let $\mathcal O_f$ be an order in $K$ with conductor $f$. Let $\chi$ be a proper character of $\operatorname{Cl}(\mathcal O_f)$ (non-principal if $f=1$). Define $$L_f(s,\chi)=\sum_{(\mathfrak a,f)=1} \chi(\mathfrak a)N(\mathfrak a)^{-s}.$$ Here the sum is over all $\mathcal O_f$-ideals prime to $f$.

Suppose that there are only two units in $O_f$. For $C\in \operatorname{Cl}(\mathcal O_f)$ and $\mathfrak b\in C^{-1},\mathfrak b \subset \mathcal O_f$, let $$\zeta(s,C)=\frac{N(\mathfrak b)^s}{2}\sum_{0\neq\gamma\in \mathfrak b}N(\gamma)^{-s}.$$

How to prove that $$L_f(s,\chi)=\sum_{C\in \operatorname{Cl}(\mathcal O_f)}\chi(C)\zeta(s,C)\quad?$$ This is proved in Curt Meyer's Die Berechnung der Klassenzahl Abelscher Körper über Quadratischen Zahlkörpern, p. 24-25, but I dont understand the argument there.

Let $K$ be an imaginary quadratic field and let $\mathcal O_f$ be an order in $K$ with conductor $f$. Let $\chi$ be a proper character of $\operatorname{Cl}(\mathcal O_f)$ (non-principal if $f=1$). Define $$L_f(s,\chi)=\sum_{(\mathfrak a,f)=1} \chi(\mathfrak a)N(\mathfrak a)^{-s}.$$ Here the sum is over all $\mathcal O_f$-ideals prime to $f$.

Suppose that there are only two units in $O_f$. For $C\in \operatorname{Cl}(\mathcal O_f)$ and $\mathfrak b\in C^{-1},\mathfrak b \subset \mathcal O_f$, let $$\zeta(s,C)=\frac{N(\mathfrak b)^s}{2}\sum_{0\neq\gamma\in \mathfrak b}N(\gamma)^{-s}.$$

How to prove that $$L_f(s,\chi)=\sum_{C\in \operatorname{Cl}(\mathcal O_f)}\chi(C)\zeta(s,C)\quad?$$ This is proved in Curt Meyer's Die Berechnung der Klassenzahl Abelscher Körper über Quadratischen Zahlkörpern, p. 24-25, but I dont understand the argument there.

The point here is that we may dispose of the condition requiring the ideals to be prime to the conductor $f$.