Extracting Dirichlet series coefficients












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Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$?






analytic-number-theory power-series dirichlet-series







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  • 2




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    Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    3 hours ago


















2












$begingroup$


Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$?






analytic-number-theory power-series dirichlet-series







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    3 hours ago
















2












2








2


1



$begingroup$


Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$?






analytic-number-theory power-series dirichlet-series







share|cite|improve this question









$endgroup$




Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$?






analytic-number-theory power-series dirichlet-series




analytic-number-theory power-series dirichlet-series






share|cite|improve this question













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asked 4 hours ago









MCHMCH

30319




30319








  • 2




    $begingroup$
    Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    3 hours ago
















  • 2




    $begingroup$
    Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
    $endgroup$
    – reuns
    3 hours ago










2




2




$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
3 hours ago






$begingroup$
Otherwise you can find the $a_n$ one by one from the asymptotic as $s to +infty$. I wonder if there is an analog of $c_n = frac{F^{(n)}(0)}{n!} = lim_{z to 0} frac{sum_{k=0}^n {k choose n} (-1)^{k-n} F(nz)}{z^n n!}$
$endgroup$
– reuns
3 hours ago












2 Answers
2






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oldest

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4












$begingroup$

Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
$$
lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
$$
IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
    $endgroup$
    – reuns
    3 hours ago



















4












$begingroup$

Even for more general Dirichlet series
$$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
there is the formula
$$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.



Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

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    active

    oldest

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    4












    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      3 hours ago
















    4












    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      3 hours ago














    4












    4








    4





    $begingroup$

    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.






    share|cite|improve this answer









    $endgroup$



    Yes. If $f(s)$ has a finite abscissa of absolute convergence $sigma_a$, then $forall sigma > sigma_a$:
    $$
    lim_{Ttoinfty} frac{1}{2T} int_{-T}^{T} f(sigma+ it)n^{it} mathrm{d}t = frac{a_n}{n^{sigma}}.
    $$
    IIRC, the proof can be found in Apostol's book on Analytic Number Theory.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 4 hours ago









    M.G.M.G.

    3,01022740




    3,01022740








    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      3 hours ago














    • 1




      $begingroup$
      I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
      $endgroup$
      – reuns
      3 hours ago








    1




    1




    $begingroup$
    I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
    $endgroup$
    – reuns
    3 hours ago




    $begingroup$
    I think it holds for $sigma > sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)int_1^infty (sum_{m le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(sigma+it)$ by $f(sigma+i.) ast frac{ k}{sqrt{2pi}} e^{-t^2 k^2/2} = sum_m a_m m^{-sigma-it} e^{- frac{ln^2m}{2 k^2}}$ to make it absolutely convergent
    $endgroup$
    – reuns
    3 hours ago











    4












    $begingroup$

    Even for more general Dirichlet series
    $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
    there is the formula
    $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
    where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.



    Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Even for more general Dirichlet series
      $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
      there is the formula
      $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
      where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.



      Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Even for more general Dirichlet series
        $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
        there is the formula
        $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
        where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.



        Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.






        share|cite|improve this answer









        $endgroup$



        Even for more general Dirichlet series
        $$f(z)=sum_{0}^infty a_n e^{-lambda_nz}$$
        there is the formula
        $$a_ne^{-lambda_nsigma}=lim_{Ttoinfty}frac{1}{T}int_{t_0}^Tf(sigma+it)e^{lambda_n it}dt,$$
        where $t_0$ is arbitrary (real) and $sigma>sigma_u$, the abscissa of uniform convergence.



        Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Alexandre EremenkoAlexandre Eremenko

        51.9k6145265




        51.9k6145265






























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