analytic number theory - orthogonality relation for quadratic Dirichlet characters

So I think I solved half of the problem. Suppose that $n$ is a perfect square. Then $chi_d(n) = 1$ unless $gcd(d,n) > 1$, in which case its $0$. So, for $gcd{d,n} = 1$, we are simply pulling out the subset of fundamental discriminants having no common divisor with $n$. To quantify the size of this subset, we must first count fundamental discriminants.

The set of fundamental discriminants consists of all square-free integers congruent to $1$ modulo $4$ (i.e. odd fundamental discriminants) and all such numbers multiplied by $-4$ and $pm 8$ (i.e. even fundamental discriminants). The odd fundamental discriminants may be counted by considering the series $$sum_{text{$d$ odd}} frac{1}{|d|^s}.$$ In fact, this is a Dirichlet series. For observe that $$sum_{text{$d$ odd}} frac{1}{|d|^s} = 1 + frac{1}{3^s} + frac{1}{5^s} + frac{1}{7^s} + cdots = prod_{p>2} left(1 + frac{1}{p^s}right) = frac{zeta(s)}{zeta(2s)} left(1 + frac{1}{2^s}right)^{-1},$$
where $$frac{zeta(s)}{zeta(2s)} = sum_{n=1}^infty frac{|mu(n)|}{n^s},$$ is the Dirichlet series which generates the square-free numbers. So, by following the definition of fundamental discriminants given above, we can count fundamental discriminants by using the Dirichlet series $$left(1 + frac{1}{4^s} + frac{2}{8^s}right) frac{zeta(s)}{zeta(2s)} left(1 +frac{1}{2^s}right)^{-1} = left(1 + frac{1}{4^s} + frac{2}{8^s}right) underbrace{prod_{p>2} left(1 + frac{1}{p^s}right)}_{l_p(s)}.$$

Now, to omit those discriminants with $gcd(d,m) > 1$, we just omit the corresponding factors in $l_p(s)$. What's missing is $$prod_{substack{p>2 \ p|m}} left(1 + frac{1}{p^s}right),$$ so the relative density (compared to all fundamental discriminants $d$) is can be quantified by the expression $$frac{1}{D} cdot prod_{substack{p>2 \ pnmid m}} left(1 + frac{1}{p^s}right) = prod_{substack{p>2 \ p|m}} left(1 + frac{1}{p^s}right)^{-1}.$$ As in the proof of the prime number theorem, the main contribution here comes from the simple pole at $s=1$ (of $zeta(s)$). In fact, $p=2$ also fits at $s=1$ since $1 + frac{1}{4} + frac{2}{8} = 1 + frac{1}{2}$. Thus, in the end we obtain $$lim_{Xrightarrow infty} frac{1}{D} sum_{0 < |d| leq X} chi_d(m) = prod_{p|m} left(1 + frac{1}{p}right)^{-1},$$ as desired.

Now, does anyone know how to explain the other case. I assume it just follows from the symmetry of the roots of unity. But can anyone validate and perhaps clarify this. Thank you.