prove $sum_{n=0}^{infty}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$












6












$begingroup$


I am seeking alternate proofs for
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Here's mine:



Recall that, for $xin(0,2)$,
$$frac1x=sum_{ngeq0}(1-x)^n$$
Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
$$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
Which gives
$$
begin{align}
int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
=&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
end{align}
$$
That last step was from the definition of the Beta function:
$$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
$$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
$$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
And with a little simplification, we arrive at
$$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!






sequences-and-series soft-question special-functions alternative-proof big-list







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    I am seeking alternate proofs for
    $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
    Here's mine:



    Recall that, for $xin(0,2)$,
    $$frac1x=sum_{ngeq0}(1-x)^n$$
    Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
    $$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
    Which gives
    $$
    begin{align}
    int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
    =&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
    =&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
    end{align}
    $$
    That last step was from the definition of the Beta function:
    $$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
    Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
    $$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
    Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
    $$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
    And with a little simplification, we arrive at
    $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
    Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!






    sequences-and-series soft-question special-functions alternative-proof big-list







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      4



      $begingroup$


      I am seeking alternate proofs for
      $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
      Here's mine:



      Recall that, for $xin(0,2)$,
      $$frac1x=sum_{ngeq0}(1-x)^n$$
      Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
      $$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
      Which gives
      $$
      begin{align}
      int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
      =&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
      =&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
      end{align}
      $$
      That last step was from the definition of the Beta function:
      $$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
      Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
      $$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
      Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
      $$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
      And with a little simplification, we arrive at
      $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
      Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!






      sequences-and-series soft-question special-functions alternative-proof big-list







      share|cite|improve this question











      $endgroup$




      I am seeking alternate proofs for
      $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
      Here's mine:



      Recall that, for $xin(0,2)$,
      $$frac1x=sum_{ngeq0}(1-x)^n$$
      Hence we have that, for $frac{1-sqrt5}2<x<frac{1+sqrt5}2$,
      $$frac1{x^2-x+1}=sum_{ngeq0}x^n(1-x)^n$$
      Which gives
      $$
      begin{align}
      int_0^1frac{mathrm dx}{x^2-x+1}=&int_0^1sum_{ngeq0}x^n(1-x)^nmathrm dx\
      =&sum_{ngeq0}int_0^1x^n(1-x)^nmathrm dx\
      =&sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}
      end{align}
      $$
      That last step was from the definition of the Beta function:
      $$B(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt=frac{Gamma(a)Gamma(b)}{Gamma(a+b)}$$
      Setting $I=int_0^1frac{mathrm dx}{x^2-x+1}$, we complete the square:
      $$I=int_0^1frac{mathrm dx}{(x-frac12)^2+frac34}$$
      Preforming the substitution $x=frac{1+3^{1/2}tan u}2$, we have
      $$I=frac{3^{1/2}}2int_{-pi/6}^{pi/6}frac{sec^2u mathrm du}{frac34+frac34tan^2u}$$
      And with a little simplification, we arrive at
      $$sum_{ngeq0}frac{Gamma^2(n+1)}{Gamma(2n+2)}=frac{2pi}{3^{3/2}}$$
      Which brings me to my questions. How else can we prove this? What other series can be evaluated using similar tricks? Have fun!






      sequences-and-series soft-question special-functions alternative-proof big-list




      sequences-and-series soft-question special-functions alternative-proof big-list






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 3:47







      clathratus

















      asked Dec 2 '18 at 21:29









      clathratusclathratus

      3,862333




      3,862333






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          A not-really-alternative approach:



          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
          and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            different enough for me! Thanks Jack (+1)
            $endgroup$
            – clathratus
            Dec 2 '18 at 21:42



















          4





          +50







          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          Hereafter, $ds{Psi}$ is the
          Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
          Gamma Function.




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
          int_{0}^{1}{1 + x over 1 + x^{3}},dd x
          \[5mm] = &
          int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
          \[5mm] = &
          {1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
          \[5mm] = &
          {1 over 6}left{bracks{%
          int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
          right.
          \[2mm] & +
          left.phantom{left{,right.}
          bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
          label{1}tag{1}
          right}
          \[5mm] = &
          {1 over 6}braces{!%
          bracks{Psipars{2 over 3} -
          Psipars{1 over 3}} +
          bracks{Psipars{5 over 6} -
          Psipars{1 over 6}}!}label{2}tag{2}
          \[5mm] = &
          {1 over 6}bracks{picotpars{pi over 3} +
          picotpars{pi over 6}} =
          {1 over 6}pars{{root{3} over 3} + root{3}}pi
          label{3}tag{3}
          \[5mm] = &
          bbx{{2root{3} over 9},pi} approx 1.2092
          end{align}




          eqref{1} is evaluated with
          $ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
          Euler Reflection Formula.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
            $endgroup$
            – clathratus
            Dec 4 '18 at 17:41






          • 1




            $begingroup$
            @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
            $endgroup$
            – Felix Marin
            Dec 4 '18 at 19:52










          • $begingroup$
            I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
            $endgroup$
            – clathratus
            Dec 8 '18 at 21:26










          • $begingroup$
            Thanks @clathratus
            $endgroup$
            – Felix Marin
            Dec 8 '18 at 23:55



















          2












          $begingroup$

          Still another way is through the Hypergeometric function.



          Since the term of the sum is
          $$
          t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
          $$
          and
          $$
          eqalign{
          & t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
          & {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
          over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
          & = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
          $$
          then
          $$
          eqalign{
          & sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
          = {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
          & = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Cool! I really like any approach that involves hypergeometric functions.
            $endgroup$
            – clathratus
            Dec 5 '18 at 0:33










          • $begingroup$
            @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
            $endgroup$
            – G Cab
            Dec 5 '18 at 1:13











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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          A not-really-alternative approach:



          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
          and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            different enough for me! Thanks Jack (+1)
            $endgroup$
            – clathratus
            Dec 2 '18 at 21:42
















          5












          $begingroup$

          A not-really-alternative approach:



          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
          and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            different enough for me! Thanks Jack (+1)
            $endgroup$
            – clathratus
            Dec 2 '18 at 21:42














          5












          5








          5





          $begingroup$

          A not-really-alternative approach:



          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
          and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$






          share|cite|improve this answer









          $endgroup$



          A not-really-alternative approach:



          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=sum_{ngeq 0}frac{1}{(2n+1)binom{2n}{n}}=sum_{ngeq 1}frac{2}{nbinom{2n}{n}} $$
          and since $sum_{ngeq 1}frac{z^{2n}}{n^2binom{2n}{n}}$ equals $2arcsin^2tfrac{z}{2}$ (by the Lagrange inversion theorem or equivalent approaches), we simply have
          $$ sum_{ngeq 0}frac{Gamma(n+1)^2}{Gamma(2n+2)}=frac{d}{dz}left.2arcsin^2tfrac{z}{2}right|_{z=1}=left.frac{4arcsin(z/2)}{sqrt{4-z^2}}right|_{z=1}=frac{2pi}{3sqrt{3}}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 21:41









          Jack D'AurizioJack D'Aurizio

          289k33281660




          289k33281660












          • $begingroup$
            different enough for me! Thanks Jack (+1)
            $endgroup$
            – clathratus
            Dec 2 '18 at 21:42


















          • $begingroup$
            different enough for me! Thanks Jack (+1)
            $endgroup$
            – clathratus
            Dec 2 '18 at 21:42
















          $begingroup$
          different enough for me! Thanks Jack (+1)
          $endgroup$
          – clathratus
          Dec 2 '18 at 21:42




          $begingroup$
          different enough for me! Thanks Jack (+1)
          $endgroup$
          – clathratus
          Dec 2 '18 at 21:42











          4





          +50







          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          Hereafter, $ds{Psi}$ is the
          Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
          Gamma Function.




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
          int_{0}^{1}{1 + x over 1 + x^{3}},dd x
          \[5mm] = &
          int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
          \[5mm] = &
          {1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
          \[5mm] = &
          {1 over 6}left{bracks{%
          int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
          right.
          \[2mm] & +
          left.phantom{left{,right.}
          bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
          label{1}tag{1}
          right}
          \[5mm] = &
          {1 over 6}braces{!%
          bracks{Psipars{2 over 3} -
          Psipars{1 over 3}} +
          bracks{Psipars{5 over 6} -
          Psipars{1 over 6}}!}label{2}tag{2}
          \[5mm] = &
          {1 over 6}bracks{picotpars{pi over 3} +
          picotpars{pi over 6}} =
          {1 over 6}pars{{root{3} over 3} + root{3}}pi
          label{3}tag{3}
          \[5mm] = &
          bbx{{2root{3} over 9},pi} approx 1.2092
          end{align}




          eqref{1} is evaluated with
          $ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
          Euler Reflection Formula.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
            $endgroup$
            – clathratus
            Dec 4 '18 at 17:41






          • 1




            $begingroup$
            @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
            $endgroup$
            – Felix Marin
            Dec 4 '18 at 19:52










          • $begingroup$
            I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
            $endgroup$
            – clathratus
            Dec 8 '18 at 21:26










          • $begingroup$
            Thanks @clathratus
            $endgroup$
            – Felix Marin
            Dec 8 '18 at 23:55
















          4





          +50







          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          Hereafter, $ds{Psi}$ is the
          Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
          Gamma Function.




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
          int_{0}^{1}{1 + x over 1 + x^{3}},dd x
          \[5mm] = &
          int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
          \[5mm] = &
          {1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
          \[5mm] = &
          {1 over 6}left{bracks{%
          int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
          right.
          \[2mm] & +
          left.phantom{left{,right.}
          bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
          label{1}tag{1}
          right}
          \[5mm] = &
          {1 over 6}braces{!%
          bracks{Psipars{2 over 3} -
          Psipars{1 over 3}} +
          bracks{Psipars{5 over 6} -
          Psipars{1 over 6}}!}label{2}tag{2}
          \[5mm] = &
          {1 over 6}bracks{picotpars{pi over 3} +
          picotpars{pi over 6}} =
          {1 over 6}pars{{root{3} over 3} + root{3}}pi
          label{3}tag{3}
          \[5mm] = &
          bbx{{2root{3} over 9},pi} approx 1.2092
          end{align}




          eqref{1} is evaluated with
          $ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
          Euler Reflection Formula.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
            $endgroup$
            – clathratus
            Dec 4 '18 at 17:41






          • 1




            $begingroup$
            @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
            $endgroup$
            – Felix Marin
            Dec 4 '18 at 19:52










          • $begingroup$
            I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
            $endgroup$
            – clathratus
            Dec 8 '18 at 21:26










          • $begingroup$
            Thanks @clathratus
            $endgroup$
            – Felix Marin
            Dec 8 '18 at 23:55














          4





          +50







          4





          +50



          4




          +50



          $begingroup$

          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          Hereafter, $ds{Psi}$ is the
          Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
          Gamma Function.




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
          int_{0}^{1}{1 + x over 1 + x^{3}},dd x
          \[5mm] = &
          int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
          \[5mm] = &
          {1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
          \[5mm] = &
          {1 over 6}left{bracks{%
          int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
          right.
          \[2mm] & +
          left.phantom{left{,right.}
          bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
          label{1}tag{1}
          right}
          \[5mm] = &
          {1 over 6}braces{!%
          bracks{Psipars{2 over 3} -
          Psipars{1 over 3}} +
          bracks{Psipars{5 over 6} -
          Psipars{1 over 6}}!}label{2}tag{2}
          \[5mm] = &
          {1 over 6}bracks{picotpars{pi over 3} +
          picotpars{pi over 6}} =
          {1 over 6}pars{{root{3} over 3} + root{3}}pi
          label{3}tag{3}
          \[5mm] = &
          bbx{{2root{3} over 9},pi} approx 1.2092
          end{align}




          eqref{1} is evaluated with
          $ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
          Euler Reflection Formula.







          share|cite|improve this answer











          $endgroup$



          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
          newcommand{dd}{mathrm{d}}
          newcommand{ds}[1]{displaystyle{#1}}
          newcommand{expo}[1]{,mathrm{e}^{#1},}
          newcommand{ic}{mathrm{i}}
          newcommand{mc}[1]{mathcal{#1}}
          newcommand{mrm}[1]{mathrm{#1}}
          newcommand{pars}[1]{left(,{#1},right)}
          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
          newcommand{verts}[1]{leftvert,{#1},rightvert}$




          Hereafter, $ds{Psi}$ is the
          Digamma Function: $ds{Psipars{z} equiv totald{lnpars{Gammapars{z}}}{z}}$ where $ds{Gamma}$ is the
          Gamma Function.




          begin{align}
          &bbox[10px,#ffd]{int_{0}^{1}{dd x over x^{2} - x + 1}} =
          int_{0}^{1}{1 + x over 1 + x^{3}},dd x
          \[5mm] = &
          int_{0}^{1}{1 + x - x^{3} - x^{4}over 1 - x^{6}},dd x
          \[5mm] = &
          {1 over 6}int_{0}^{1}{x^{-5/6} + x^{-2/3} - x^{-1/3} - x^{-1/6} over 1 - x},dd x
          \[5mm] = &
          {1 over 6}left{bracks{%
          int_{0}^{1}{1 - x^{-1/3} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-2/3} over 1 - x},dd x}
          right.
          \[2mm] & +
          left.phantom{left{,right.}
          bracks{int_{0}^{1}{1 - x^{-1/6} over 1 - x},dd x -
          int_{0}^{1}{1 - x^{-5/6} over 1 - x},dd x}
          label{1}tag{1}
          right}
          \[5mm] = &
          {1 over 6}braces{!%
          bracks{Psipars{2 over 3} -
          Psipars{1 over 3}} +
          bracks{Psipars{5 over 6} -
          Psipars{1 over 6}}!}label{2}tag{2}
          \[5mm] = &
          {1 over 6}bracks{picotpars{pi over 3} +
          picotpars{pi over 6}} =
          {1 over 6}pars{{root{3} over 3} + root{3}}pi
          label{3}tag{3}
          \[5mm] = &
          bbx{{2root{3} over 9},pi} approx 1.2092
          end{align}




          eqref{1} is evaluated with
          $ds{mathbf{color{black}{6.3.22}}}$ A & S Identity. In eqref{2} and eqref{3}, I used the
          Euler Reflection Formula.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 19:56

























          answered Dec 4 '18 at 5:57









          Felix MarinFelix Marin

          67.5k7107141




          67.5k7107141












          • $begingroup$
            Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
            $endgroup$
            – clathratus
            Dec 4 '18 at 17:41






          • 1




            $begingroup$
            @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
            $endgroup$
            – Felix Marin
            Dec 4 '18 at 19:52










          • $begingroup$
            I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
            $endgroup$
            – clathratus
            Dec 8 '18 at 21:26










          • $begingroup$
            Thanks @clathratus
            $endgroup$
            – Felix Marin
            Dec 8 '18 at 23:55


















          • $begingroup$
            Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
            $endgroup$
            – clathratus
            Dec 4 '18 at 17:41






          • 1




            $begingroup$
            @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
            $endgroup$
            – Felix Marin
            Dec 4 '18 at 19:52










          • $begingroup$
            I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
            $endgroup$
            – clathratus
            Dec 8 '18 at 21:26










          • $begingroup$
            Thanks @clathratus
            $endgroup$
            – Felix Marin
            Dec 8 '18 at 23:55
















          $begingroup$
          Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
          $endgroup$
          – clathratus
          Dec 4 '18 at 17:41




          $begingroup$
          Nice! Thanks for another one of your high quality answers, Felix. Just for clarification, $$Psi(z)=frac{Gamma'(z)}{Gamma(z)}?$$
          $endgroup$
          – clathratus
          Dec 4 '18 at 17:41




          1




          1




          $begingroup$
          @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
          $endgroup$
          – Felix Marin
          Dec 4 '18 at 19:52




          $begingroup$
          @clathratus Yes. I just added some comments to my answers which clarifies it. Thanks.
          $endgroup$
          – Felix Marin
          Dec 4 '18 at 19:52












          $begingroup$
          I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
          $endgroup$
          – clathratus
          Dec 8 '18 at 21:26




          $begingroup$
          I'd like to award a bounty to this answer, as it provides the most alternative, integration-related proof. Excellent work Felix.
          $endgroup$
          – clathratus
          Dec 8 '18 at 21:26












          $begingroup$
          Thanks @clathratus
          $endgroup$
          – Felix Marin
          Dec 8 '18 at 23:55




          $begingroup$
          Thanks @clathratus
          $endgroup$
          – Felix Marin
          Dec 8 '18 at 23:55











          2












          $begingroup$

          Still another way is through the Hypergeometric function.



          Since the term of the sum is
          $$
          t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
          $$
          and
          $$
          eqalign{
          & t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
          & {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
          over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
          & = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
          $$
          then
          $$
          eqalign{
          & sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
          = {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
          & = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Cool! I really like any approach that involves hypergeometric functions.
            $endgroup$
            – clathratus
            Dec 5 '18 at 0:33










          • $begingroup$
            @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
            $endgroup$
            – G Cab
            Dec 5 '18 at 1:13
















          2












          $begingroup$

          Still another way is through the Hypergeometric function.



          Since the term of the sum is
          $$
          t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
          $$
          and
          $$
          eqalign{
          & t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
          & {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
          over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
          & = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
          $$
          then
          $$
          eqalign{
          & sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
          = {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
          & = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Cool! I really like any approach that involves hypergeometric functions.
            $endgroup$
            – clathratus
            Dec 5 '18 at 0:33










          • $begingroup$
            @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
            $endgroup$
            – G Cab
            Dec 5 '18 at 1:13














          2












          2








          2





          $begingroup$

          Still another way is through the Hypergeometric function.



          Since the term of the sum is
          $$
          t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
          $$
          and
          $$
          eqalign{
          & t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
          & {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
          over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
          & = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
          $$
          then
          $$
          eqalign{
          & sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
          = {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
          & = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
          $$






          share|cite|improve this answer









          $endgroup$



          Still another way is through the Hypergeometric function.



          Since the term of the sum is
          $$
          t_{,n} = {{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2n + 2} right)}}
          $$
          and
          $$
          eqalign{
          & t_{,0} = {{Gamma left( 1 right)^{,2} } over {Gamma left( 2 right)}} = 1 cr
          & {{t_{,n + 1} } over {t_{,n} }} = {{Gamma left( {n + 2} right)^{,2} } over {Gamma left( {2n + 4} right)}}{{Gamma left( {2n + 2} right)}
          over {Gamma left( {n + 1} right)^{,2} }} = {{left( {n + 1} right)^{,2} } over {left( {2n + 3} right)left( {2n + 2} right)}} cr
          & = {{left( {n + 1} right)left( {n + 1} right)} over {left( {n + 3/2} right)}}{{1/4} over {left( {n + 1} right)}} cr}
          $$
          then
          $$
          eqalign{
          & sumlimits_{n = 0}^infty {{{Gamma left( {n + 1} right)^{,2} } over {Gamma left( {2left( {n + 1} right)} right)}}}
          = {}_2F_{,1} left( {left. {matrix{ {1,;1} cr {3/2} cr } ;} right|;1/4} right) = cr
          & = {{arcsin left( {sqrt {1/4} } right)} over {sqrt {1 - 1/4} sqrt {1/4} }} = {{4pi } over {6sqrt 3 }} = {{2pi } over {3sqrt 3 }} cr}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 23:19









          G CabG Cab

          18.6k31238




          18.6k31238












          • $begingroup$
            Cool! I really like any approach that involves hypergeometric functions.
            $endgroup$
            – clathratus
            Dec 5 '18 at 0:33










          • $begingroup$
            @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
            $endgroup$
            – G Cab
            Dec 5 '18 at 1:13


















          • $begingroup$
            Cool! I really like any approach that involves hypergeometric functions.
            $endgroup$
            – clathratus
            Dec 5 '18 at 0:33










          • $begingroup$
            @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
            $endgroup$
            – G Cab
            Dec 5 '18 at 1:13
















          $begingroup$
          Cool! I really like any approach that involves hypergeometric functions.
          $endgroup$
          – clathratus
          Dec 5 '18 at 0:33




          $begingroup$
          Cool! I really like any approach that involves hypergeometric functions.
          $endgroup$
          – clathratus
          Dec 5 '18 at 0:33












          $begingroup$
          @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
          $endgroup$
          – G Cab
          Dec 5 '18 at 1:13




          $begingroup$
          @clathratus: glad of the appreciation, but I am aware of the drawback that the HG does not have a straight translation into the final result.
          $endgroup$
          – G Cab
          Dec 5 '18 at 1:13


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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