Find the residue of $frac{zeta'(1+s)}{zeta(1+s)}frac{x^s}{s}$ at $s=0$












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$begingroup$


Let, $displaystyle f(s)=frac{zeta'(1+s)}{zeta(1+s)}frac{x^s}{s}$. Prove that $Res(f,s=0)=A-log x$ , for some constant $A$.



At $s=0$ , $zeta(s)$ has a pole of order $1$ and $zeta'(s)$ has a pole of order $2$. So, $frac{zeta'(1+s)}{zeta(1+s)}$ has a simple pole at $s=0$. Again , $x^s/s$ has a simple pole at $s=0$. So $f$ has a pole of order $2$ at $s=0$. So the residue is given by



$$lim_{sto 0}frac{d}{ds}left{sx^s.frac{zeta'(1+s)}{zeta(1+s)}right}.$$



I've stuck here ! How can I calculate this to prove the result ?






complex-analysis riemann-zeta zeta-functions







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  • 1




    $begingroup$
    Hint: The answer should be that A is Euler–Mascheroni constant here.
    $endgroup$
    – Nanayajitzuki
    Dec 9 '18 at 8:06










  • $begingroup$
    $frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
    $endgroup$
    – reuns
    Dec 9 '18 at 8:09


















1












$begingroup$


Let, $displaystyle f(s)=frac{zeta'(1+s)}{zeta(1+s)}frac{x^s}{s}$. Prove that $Res(f,s=0)=A-log x$ , for some constant $A$.



At $s=0$ , $zeta(s)$ has a pole of order $1$ and $zeta'(s)$ has a pole of order $2$. So, $frac{zeta'(1+s)}{zeta(1+s)}$ has a simple pole at $s=0$. Again , $x^s/s$ has a simple pole at $s=0$. So $f$ has a pole of order $2$ at $s=0$. So the residue is given by



$$lim_{sto 0}frac{d}{ds}left{sx^s.frac{zeta'(1+s)}{zeta(1+s)}right}.$$



I've stuck here ! How can I calculate this to prove the result ?






complex-analysis riemann-zeta zeta-functions







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: The answer should be that A is Euler–Mascheroni constant here.
    $endgroup$
    – Nanayajitzuki
    Dec 9 '18 at 8:06










  • $begingroup$
    $frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
    $endgroup$
    – reuns
    Dec 9 '18 at 8:09
















1












1








1





$begingroup$


Let, $displaystyle f(s)=frac{zeta'(1+s)}{zeta(1+s)}frac{x^s}{s}$. Prove that $Res(f,s=0)=A-log x$ , for some constant $A$.



At $s=0$ , $zeta(s)$ has a pole of order $1$ and $zeta'(s)$ has a pole of order $2$. So, $frac{zeta'(1+s)}{zeta(1+s)}$ has a simple pole at $s=0$. Again , $x^s/s$ has a simple pole at $s=0$. So $f$ has a pole of order $2$ at $s=0$. So the residue is given by



$$lim_{sto 0}frac{d}{ds}left{sx^s.frac{zeta'(1+s)}{zeta(1+s)}right}.$$



I've stuck here ! How can I calculate this to prove the result ?






complex-analysis riemann-zeta zeta-functions







share|cite|improve this question









$endgroup$




Let, $displaystyle f(s)=frac{zeta'(1+s)}{zeta(1+s)}frac{x^s}{s}$. Prove that $Res(f,s=0)=A-log x$ , for some constant $A$.



At $s=0$ , $zeta(s)$ has a pole of order $1$ and $zeta'(s)$ has a pole of order $2$. So, $frac{zeta'(1+s)}{zeta(1+s)}$ has a simple pole at $s=0$. Again , $x^s/s$ has a simple pole at $s=0$. So $f$ has a pole of order $2$ at $s=0$. So the residue is given by



$$lim_{sto 0}frac{d}{ds}left{sx^s.frac{zeta'(1+s)}{zeta(1+s)}right}.$$



I've stuck here ! How can I calculate this to prove the result ?






complex-analysis riemann-zeta zeta-functions




complex-analysis riemann-zeta zeta-functions






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asked Dec 9 '18 at 7:30









TopoTopo

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323314








  • 1




    $begingroup$
    Hint: The answer should be that A is Euler–Mascheroni constant here.
    $endgroup$
    – Nanayajitzuki
    Dec 9 '18 at 8:06










  • $begingroup$
    $frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
    $endgroup$
    – reuns
    Dec 9 '18 at 8:09
















  • 1




    $begingroup$
    Hint: The answer should be that A is Euler–Mascheroni constant here.
    $endgroup$
    – Nanayajitzuki
    Dec 9 '18 at 8:06










  • $begingroup$
    $frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
    $endgroup$
    – reuns
    Dec 9 '18 at 8:09










1




1




$begingroup$
Hint: The answer should be that A is Euler–Mascheroni constant here.
$endgroup$
– Nanayajitzuki
Dec 9 '18 at 8:06




$begingroup$
Hint: The answer should be that A is Euler–Mascheroni constant here.
$endgroup$
– Nanayajitzuki
Dec 9 '18 at 8:06












$begingroup$
$frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
$endgroup$
– reuns
Dec 9 '18 at 8:09






$begingroup$
$frac{x^s}{s} = frac{1}{s}+log x+O(s)$, do the same with $zeta'/zeta(1+s)$ and pick the coef of $1/s$ (which appears when integrating $int_{|s|=epsilon}$)
$endgroup$
– reuns
Dec 9 '18 at 8:09












2 Answers
2






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oldest

votes


















1












$begingroup$

You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $zeta(s)$



$$displaystyle zeta (s)={frac{1}{s-1}}+sum _{n=0}^{infty}{frac{(-1)^{n}}{n!}}gamma _{n}(s-1)^{n}$$



and you take the first and constant item



$$displaystyle zeta (s+1)={frac{1}{s}}+gamma_{0}+O(s)$$



then you can find



$$displaystyle begin{align}
frac{zeta'(1+s)}{zeta(1+s)}&=frac{dln(zeta(1+s))}{ds}\&=frac{d}{ds}left(ln(1+gamma_{0}s+O(s^2))-ln sright)=frac{d}{ds}left(gamma_{0}s+O(s^2)-ln sright)\&={-frac{1}{s}}+gamma_{0}+O(s)
end{align}$$



notice you have



$$displaystyle frac{x^s}{s}=frac{1}{s}+ln x+O(s)$$



so you can easily get



$$displaystyle f(s)=-frac{1}{s^2}+frac{gamma_{0}-ln x}{s}+O(1)$$



here $gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Due to the integral representation for the $zeta$ function, in a neighbourhood of $s=0$ we have
    $$ zeta(s+1)=frac{1}{s}+gamma+(gamma_1+o(1))s = frac{1}{s}left[1+gamma s+O(1)sright]tag{1} $$
    hence by applying $frac{d}{ds}log(cdot)$ to both sides
    $$ frac{zeta'(s+1)}{zeta(s+1)} = -frac{1}{s}+gamma+O(1)s tag{2} $$
    while
    $$ frac{x^s}{s} = frac{1}{s}exp(slog x)=frac{1}{s}+log x+O(1)slog^2(x)tag{3}$$
    hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $frac{zeta'(s+1)}{zeta(s+1)}cdotfrac{x^s}{s}$ with residue $gamma-log x$.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $zeta(s)$



      $$displaystyle zeta (s)={frac{1}{s-1}}+sum _{n=0}^{infty}{frac{(-1)^{n}}{n!}}gamma _{n}(s-1)^{n}$$



      and you take the first and constant item



      $$displaystyle zeta (s+1)={frac{1}{s}}+gamma_{0}+O(s)$$



      then you can find



      $$displaystyle begin{align}
      frac{zeta'(1+s)}{zeta(1+s)}&=frac{dln(zeta(1+s))}{ds}\&=frac{d}{ds}left(ln(1+gamma_{0}s+O(s^2))-ln sright)=frac{d}{ds}left(gamma_{0}s+O(s^2)-ln sright)\&={-frac{1}{s}}+gamma_{0}+O(s)
      end{align}$$



      notice you have



      $$displaystyle frac{x^s}{s}=frac{1}{s}+ln x+O(s)$$



      so you can easily get



      $$displaystyle f(s)=-frac{1}{s^2}+frac{gamma_{0}-ln x}{s}+O(1)$$



      here $gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $zeta(s)$



        $$displaystyle zeta (s)={frac{1}{s-1}}+sum _{n=0}^{infty}{frac{(-1)^{n}}{n!}}gamma _{n}(s-1)^{n}$$



        and you take the first and constant item



        $$displaystyle zeta (s+1)={frac{1}{s}}+gamma_{0}+O(s)$$



        then you can find



        $$displaystyle begin{align}
        frac{zeta'(1+s)}{zeta(1+s)}&=frac{dln(zeta(1+s))}{ds}\&=frac{d}{ds}left(ln(1+gamma_{0}s+O(s^2))-ln sright)=frac{d}{ds}left(gamma_{0}s+O(s^2)-ln sright)\&={-frac{1}{s}}+gamma_{0}+O(s)
        end{align}$$



        notice you have



        $$displaystyle frac{x^s}{s}=frac{1}{s}+ln x+O(s)$$



        so you can easily get



        $$displaystyle f(s)=-frac{1}{s^2}+frac{gamma_{0}-ln x}{s}+O(1)$$



        here $gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $zeta(s)$



          $$displaystyle zeta (s)={frac{1}{s-1}}+sum _{n=0}^{infty}{frac{(-1)^{n}}{n!}}gamma _{n}(s-1)^{n}$$



          and you take the first and constant item



          $$displaystyle zeta (s+1)={frac{1}{s}}+gamma_{0}+O(s)$$



          then you can find



          $$displaystyle begin{align}
          frac{zeta'(1+s)}{zeta(1+s)}&=frac{dln(zeta(1+s))}{ds}\&=frac{d}{ds}left(ln(1+gamma_{0}s+O(s^2))-ln sright)=frac{d}{ds}left(gamma_{0}s+O(s^2)-ln sright)\&={-frac{1}{s}}+gamma_{0}+O(s)
          end{align}$$



          notice you have



          $$displaystyle frac{x^s}{s}=frac{1}{s}+ln x+O(s)$$



          so you can easily get



          $$displaystyle f(s)=-frac{1}{s^2}+frac{gamma_{0}-ln x}{s}+O(1)$$



          here $gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.






          share|cite|improve this answer











          $endgroup$



          You could get an easy access to answer if you already know the Stieltjes constants which appears in the Laurent series expansion of $zeta(s)$



          $$displaystyle zeta (s)={frac{1}{s-1}}+sum _{n=0}^{infty}{frac{(-1)^{n}}{n!}}gamma _{n}(s-1)^{n}$$



          and you take the first and constant item



          $$displaystyle zeta (s+1)={frac{1}{s}}+gamma_{0}+O(s)$$



          then you can find



          $$displaystyle begin{align}
          frac{zeta'(1+s)}{zeta(1+s)}&=frac{dln(zeta(1+s))}{ds}\&=frac{d}{ds}left(ln(1+gamma_{0}s+O(s^2))-ln sright)=frac{d}{ds}left(gamma_{0}s+O(s^2)-ln sright)\&={-frac{1}{s}}+gamma_{0}+O(s)
          end{align}$$



          notice you have



          $$displaystyle frac{x^s}{s}=frac{1}{s}+ln x+O(s)$$



          so you can easily get



          $$displaystyle f(s)=-frac{1}{s^2}+frac{gamma_{0}-ln x}{s}+O(1)$$



          here $gamma_{0}$ is Euler–Mascheroni constant, and the simply proof on the constant appears in the series can see another question like this one.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 11:48

























          answered Dec 9 '18 at 11:40









          NanayajitzukiNanayajitzuki

          3185




          3185























              1












              $begingroup$

              Due to the integral representation for the $zeta$ function, in a neighbourhood of $s=0$ we have
              $$ zeta(s+1)=frac{1}{s}+gamma+(gamma_1+o(1))s = frac{1}{s}left[1+gamma s+O(1)sright]tag{1} $$
              hence by applying $frac{d}{ds}log(cdot)$ to both sides
              $$ frac{zeta'(s+1)}{zeta(s+1)} = -frac{1}{s}+gamma+O(1)s tag{2} $$
              while
              $$ frac{x^s}{s} = frac{1}{s}exp(slog x)=frac{1}{s}+log x+O(1)slog^2(x)tag{3}$$
              hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $frac{zeta'(s+1)}{zeta(s+1)}cdotfrac{x^s}{s}$ with residue $gamma-log x$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Due to the integral representation for the $zeta$ function, in a neighbourhood of $s=0$ we have
                $$ zeta(s+1)=frac{1}{s}+gamma+(gamma_1+o(1))s = frac{1}{s}left[1+gamma s+O(1)sright]tag{1} $$
                hence by applying $frac{d}{ds}log(cdot)$ to both sides
                $$ frac{zeta'(s+1)}{zeta(s+1)} = -frac{1}{s}+gamma+O(1)s tag{2} $$
                while
                $$ frac{x^s}{s} = frac{1}{s}exp(slog x)=frac{1}{s}+log x+O(1)slog^2(x)tag{3}$$
                hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $frac{zeta'(s+1)}{zeta(s+1)}cdotfrac{x^s}{s}$ with residue $gamma-log x$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Due to the integral representation for the $zeta$ function, in a neighbourhood of $s=0$ we have
                  $$ zeta(s+1)=frac{1}{s}+gamma+(gamma_1+o(1))s = frac{1}{s}left[1+gamma s+O(1)sright]tag{1} $$
                  hence by applying $frac{d}{ds}log(cdot)$ to both sides
                  $$ frac{zeta'(s+1)}{zeta(s+1)} = -frac{1}{s}+gamma+O(1)s tag{2} $$
                  while
                  $$ frac{x^s}{s} = frac{1}{s}exp(slog x)=frac{1}{s}+log x+O(1)slog^2(x)tag{3}$$
                  hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $frac{zeta'(s+1)}{zeta(s+1)}cdotfrac{x^s}{s}$ with residue $gamma-log x$.






                  share|cite|improve this answer









                  $endgroup$



                  Due to the integral representation for the $zeta$ function, in a neighbourhood of $s=0$ we have
                  $$ zeta(s+1)=frac{1}{s}+gamma+(gamma_1+o(1))s = frac{1}{s}left[1+gamma s+O(1)sright]tag{1} $$
                  hence by applying $frac{d}{ds}log(cdot)$ to both sides
                  $$ frac{zeta'(s+1)}{zeta(s+1)} = -frac{1}{s}+gamma+O(1)s tag{2} $$
                  while
                  $$ frac{x^s}{s} = frac{1}{s}exp(slog x)=frac{1}{s}+log x+O(1)slog^2(x)tag{3}$$
                  hence by multiplying $(2)$ and $(3)$ we get that $s=0$ is a double pole of $frac{zeta'(s+1)}{zeta(s+1)}cdotfrac{x^s}{s}$ with residue $gamma-log x$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 11:31









                  Jack D'AurizioJack D'Aurizio

                  290k33282662




                  290k33282662






























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