Evaluating Dirichlet $L$-functions at $s=1$












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$begingroup$


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.










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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.










      share|cite|improve this question









      $endgroup$




      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.







      number-theory closed-form l-functions






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      asked Jan 15 at 11:27









      ribbcastleribbcastle

      594




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          2 Answers
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          $begingroup$

          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
          $$
          L(1, chi) = prod_{p | m, p nmid m'} (1-chi'(p)p^{-1})
          L(1,chi').
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Should the condition on the extra product not be $pmid m, pnotmid m'$?
            $endgroup$
            – 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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
            $endgroup$
            – ribbcastle
            Jan 19 at 14:58





















          1












          $begingroup$

          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 $$widehat{chi}(k) = overline{chi(k)} widehat{chi}(1), qquad |widehat{chi}(1)|^2 = q$$






          share|cite|improve this answer











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            2 Answers
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            2 Answers
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            active

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            $begingroup$

            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
            $$
            L(1, chi) = prod_{p | m, p nmid m'} (1-chi'(p)p^{-1})
            L(1,chi').
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Should the condition on the extra product not be $pmid m, pnotmid m'$?
              $endgroup$
              – 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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
              $endgroup$
              – ribbcastle
              Jan 19 at 14:58


















            2












            $begingroup$

            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
            $$
            L(1, chi) = prod_{p | m, p nmid m'} (1-chi'(p)p^{-1})
            L(1,chi').
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Should the condition on the extra product not be $pmid m, pnotmid m'$?
              $endgroup$
              – 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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
              $endgroup$
              – ribbcastle
              Jan 19 at 14:58
















            2












            2








            2





            $begingroup$

            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
            $$
            L(1, chi) = prod_{p | m, p nmid m'} (1-chi'(p)p^{-1})
            L(1,chi').
            $$






            share|cite|improve this answer









            $endgroup$



            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
            $$
            L(1, chi) = prod_{p | m, p nmid m'} (1-chi'(p)p^{-1})
            L(1,chi').
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 15 at 21:24









            KimballKimball

            2,0551029




            2,0551029












            • $begingroup$
              Should the condition on the extra product not be $pmid m, pnotmid m'$?
              $endgroup$
              – 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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
              $endgroup$
              – ribbcastle
              Jan 19 at 14:58




















            • $begingroup$
              Should the condition on the extra product not be $pmid m, pnotmid m'$?
              $endgroup$
              – 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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
              $endgroup$
              – ribbcastle
              Jan 19 at 14:58


















            $begingroup$
            Should the condition on the extra product not be $pmid m, pnotmid m'$?
            $endgroup$
            – ribbcastle
            Jan 18 at 12:13




            $begingroup$
            Should the condition on the extra product not be $pmid m, pnotmid m'$?
            $endgroup$
            – 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




            $begingroup$
            @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, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
            $endgroup$
            – ribbcastle
            Jan 19 at 14:58






            $begingroup$
            apologies! the resolution on my screen is not great! I copied down an incorrect version, $pmid m$, $pmid m'$, and I guess just stopped being able to see what you'd actually written.
            $endgroup$
            – ribbcastle
            Jan 19 at 14:58













            1












            $begingroup$

            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 $$widehat{chi}(k) = overline{chi(k)} widehat{chi}(1), qquad |widehat{chi}(1)|^2 = q$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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 $$widehat{chi}(k) = overline{chi(k)} widehat{chi}(1), qquad |widehat{chi}(1)|^2 = q$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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 $$widehat{chi}(k) = overline{chi(k)} widehat{chi}(1), qquad |widehat{chi}(1)|^2 = q$$






                share|cite|improve this answer











                $endgroup$



                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 $$widehat{chi}(k) = overline{chi(k)} widehat{chi}(1), qquad |widehat{chi}(1)|^2 = q$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 16 at 3:02

























                answered Jan 16 at 2:52









                reunsreuns

                21k21354




                21k21354






























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