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I'm trying to find references on general methods for evaluating Dirichlet $L$-functions at $s=1$, but it's proving a little harder to google than I'd hoped. Specifically I'm looking for any books or papers that go through methods for solving the following question: suppose I have a (possibly primitive) Dirichlet character $\chi$ of conductor $q$, what is the value of $L(1,\chi)$? Any insights anyone has about this that aren't a book or paper would also be very much appreciated.

This question has a good amount of detail on how to approach the problem, but it only covers odd characters. The Wikipedia page for the analytic class number formula also has this section, which gives closed forms for $L$-functions at $s=1$, but this is only for primitive characters with prime conductors (and possibly only for quadratic characters?) This is as much as I could find, and as for the more general cases I'm at a bit of a loss.

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For primitive characters $\chi$ mod $m$, odd or even, a formula is given in Prop 3.4.8 of my Number Theory II notes.

If I have a character $\chi$ mod $m$ which is not primitive, but say comes from the primitive character $\chi'$ mod $m'$, then the values $\chi(p)$ and $\chi'(p)$ are only different if $p | m$ but $p \nmid m'$ (in which case $\chi(p) = 0$). So by the Euler product, we have $$ L(s, \chi) = \prod_p (1-\chi(p)p^{-s})^{-1} = \prod_{p | m, p \nmid m'} (1-\chi'(p)p^{-s}) \times L(s, \chi'). $$

Thus one can use the formula in the primitive case to compute $L(1,\chi')$ and consequently $$ L(1, \chi) = \prod_{p | m, p \nmid m'} (1-\chi'(p)p^{-1}) L(1,\chi'). $$

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  • $\begingroup$ Should the condition on the extra product not be $p\mid m, p\not\mid m'$? $\endgroup$ – ribbcastle Jan 18 at 12:13
  • $\begingroup$ @ribbcastle I'm confused about your condition---that's the same condition I wrote on the product. $\endgroup$ – Kimball Jan 18 at 14:37
  • $\begingroup$ apologies! the resolution on my screen is not great! I copied down an incorrect version, $p\mid m$, $p\mid m'$, and I guess just stopped being able to see what you'd actually written. $\endgroup$ – ribbcastle Jan 19 at 14:58
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Let $$f(x ) =\sum_{n=1}^\infty \frac{\chi(n)}{n} e^{-nx}$$ For $\chi(n)$ a non-trivial Dirichlet character modulo $q$ then $$\Gamma(s)L(s+1,\chi) = \int_0^\infty f(x)x^{s-1}dx$$ converges and is analytic for $\Re(s) > 0$.

$$\Gamma(s)L(s+1,\chi)-f(0) \Gamma(s) = \int_0^\infty (f(x)-f(0) e^{-x})x^{s-1}dx$$ converges and is analytic for $\Re(s) > -1$.

Since $\Gamma(s)$ has a pole at $s=0$ it means $$L(1,\chi) = f(0) $$ The discrete Fourier transform gives $$\chi(n) = \frac{1}{q} \sum_{k=1}^{q} \widehat{\chi}(k) e^{2i \pi nk/q}, \qquad \widehat{\chi}(k)=\sum_{n=1}^q \chi(n) e^{-2i \pi nk/q}, \qquad \widehat{\chi}(q)=0$$ Whence $$f(0)= \lim_{x \to 0}\frac{1}{q} \sum_{k=1}^{q-1} \widehat{\chi}(k) \sum_{n=1}^\infty \frac{e^{2i \pi nk/q}}{n} e^{-nx}=-\frac{1}{q} \sum_{k=1}^{q-1} \widehat{\chi}(k) \lim_{x \to 0} \log(1-e^{-x+2i \pi k/q})\\= -\frac{1}{q} \sum_{k=1}^{q-1} \widehat{\chi}(k) \log(1-e^{2i \pi k/q}) $$

Iff $\chi$ is a primitive character then $$\widehat{\chi}(k) = \overline{\chi(k)} \ \widehat{\chi}(1), \qquad |\widehat{\chi}(1)|^2 = q$$

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