Evaluate $sum_{n=1}^{infty}frac{psi^{''}(n)}{2n-1}$, where $psi^{''}(n)$ is 2nd derivative of digamma...












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Does the following sum have a closed form?
$$sum_{n=1}^{infty}frac{psi^{"}(n)}{2n-1},$$ where $psi^{"}(n)$ is 2nd derivative of digamma function.






sequences-and-series summation digamma-function







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    1












    $begingroup$


    Does the following sum have a closed form?
    $$sum_{n=1}^{infty}frac{psi^{"}(n)}{2n-1},$$ where $psi^{"}(n)$ is 2nd derivative of digamma function.






    sequences-and-series summation digamma-function







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Does the following sum have a closed form?
      $$sum_{n=1}^{infty}frac{psi^{"}(n)}{2n-1},$$ where $psi^{"}(n)$ is 2nd derivative of digamma function.






      sequences-and-series summation digamma-function







      share|cite|improve this question











      $endgroup$




      Does the following sum have a closed form?
      $$sum_{n=1}^{infty}frac{psi^{"}(n)}{2n-1},$$ where $psi^{"}(n)$ is 2nd derivative of digamma function.






      sequences-and-series summation digamma-function




      sequences-and-series summation digamma-function






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Jan 27 at 13:05









      p4sch

      5,460318




      5,460318










      asked Jan 26 at 11:40









      ben tenysonben tenyson

      414




      414






















          1 Answer
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          $begingroup$

          The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
          $$psi^{(m)}(x) = (-1)^{m+1} int_0^infty frac{t^m}{1-e^{-t}} e^{-xt} , mathrm{d} t.$$
          Using the monotone convergence theorem, we get
          $$label{1} tag{1}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} sum_{n=1}^infty frac{e^{-nt}}{2n-1} , mathrm{d}t. $$
          Since
          $$sum_{n=1}^infty e^{-(n-1/2)t} = frac{e^{t/2}}{e^t-1},$$
          we get by integration
          $$ sum_{n=1}^infty frac{e^{-nt}}{2n-1} = frac{e^{-t/2}}{2} int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s.$$
          But we have
          $$int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s = 2 int_{e^{t/2}}^infty frac{1}{x^2-1} , mathrm{d} x = 2 mathrm{arcoth}(e^{t/2}) $$
          and therefore
          $$sum_{n=1}^infty frac{e^{-nt}}{2n-1} = e^{-t/2} mathrm{arcoth}(e^{t/2}).$$
          Note that we have proven that $$mathrm{arcoth}(x) = sum_{n=1}^infty frac{x^{-(2n-1)}}{2n-1} quad text{for} quad |x| >1.$$




          Thus, we can rewrite eqref{1} by
          $$tag{2}label{2}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t.$$




          Changing variables we see that
          begin{align}
          - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t = - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x.
          end{align}
          Using $mathrm{artanh}(1/x) = mathrm{coth}(x)$ we can rewrite the integral also by
          $$ - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x= 8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y.$$




          With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.




          Using partial integration in the last line we get that
          begin{align}
          8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y &= left.-4 mathrm{artanh}(y)^2 ln(y)^2 right|_{y=0}^1 + 8 int_0mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y \
          &= 8 int_0^1mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y.
          end{align}
          Since $$mathrm{artanh}(y) = frac{1}{2} left( ln(y+1)-ln(1-y) right) $$
          we get that eqref{2} is equal to
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y - 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y + 2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y $$
          The remaining integrals are evaluated here, here and here. In fact, we have
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y = - zeta(4) = - frac{pi^4}{90} $$
          and
          $$- 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y = frac{3}{40} pi^4 - 7 log(2) zeta(3) + frac{pi^2 log(2)^2}{3}- frac{log(2)^4}{3} - 8 mathrm{Li}_4(1/2)$$
          and also
          $$2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y = frac{pi^2}{12} - 8 mathrm{Li}_4(1/2) -7 log(2) zeta(3)+ frac{pi^2 log(2)^2}{3} - frac{ln(2)^4}{3} $$




          Finally, we see that eqref{2} can be written as
          $$frac{53}{360} pi^4 - 14 log(2) zeta(3) + frac{2}{3} [pi^2 -log(2)^2] log(2)^2 - 16 mathrm{Li}_4(1/2).$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
            $endgroup$
            – ben tenyson
            Jan 26 at 14:51








          • 4




            $begingroup$
            The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
            $endgroup$
            – ComplexYetTrivial
            Jan 26 at 15:04










          • $begingroup$
            The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
            $endgroup$
            – p4sch
            Jan 27 at 12:59












          • $begingroup$
            @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
            $endgroup$
            – ben tenyson
            Jan 27 at 13:26













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          active

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          3












          $begingroup$

          The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
          $$psi^{(m)}(x) = (-1)^{m+1} int_0^infty frac{t^m}{1-e^{-t}} e^{-xt} , mathrm{d} t.$$
          Using the monotone convergence theorem, we get
          $$label{1} tag{1}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} sum_{n=1}^infty frac{e^{-nt}}{2n-1} , mathrm{d}t. $$
          Since
          $$sum_{n=1}^infty e^{-(n-1/2)t} = frac{e^{t/2}}{e^t-1},$$
          we get by integration
          $$ sum_{n=1}^infty frac{e^{-nt}}{2n-1} = frac{e^{-t/2}}{2} int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s.$$
          But we have
          $$int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s = 2 int_{e^{t/2}}^infty frac{1}{x^2-1} , mathrm{d} x = 2 mathrm{arcoth}(e^{t/2}) $$
          and therefore
          $$sum_{n=1}^infty frac{e^{-nt}}{2n-1} = e^{-t/2} mathrm{arcoth}(e^{t/2}).$$
          Note that we have proven that $$mathrm{arcoth}(x) = sum_{n=1}^infty frac{x^{-(2n-1)}}{2n-1} quad text{for} quad |x| >1.$$




          Thus, we can rewrite eqref{1} by
          $$tag{2}label{2}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t.$$




          Changing variables we see that
          begin{align}
          - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t = - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x.
          end{align}
          Using $mathrm{artanh}(1/x) = mathrm{coth}(x)$ we can rewrite the integral also by
          $$ - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x= 8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y.$$




          With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.




          Using partial integration in the last line we get that
          begin{align}
          8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y &= left.-4 mathrm{artanh}(y)^2 ln(y)^2 right|_{y=0}^1 + 8 int_0mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y \
          &= 8 int_0^1mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y.
          end{align}
          Since $$mathrm{artanh}(y) = frac{1}{2} left( ln(y+1)-ln(1-y) right) $$
          we get that eqref{2} is equal to
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y - 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y + 2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y $$
          The remaining integrals are evaluated here, here and here. In fact, we have
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y = - zeta(4) = - frac{pi^4}{90} $$
          and
          $$- 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y = frac{3}{40} pi^4 - 7 log(2) zeta(3) + frac{pi^2 log(2)^2}{3}- frac{log(2)^4}{3} - 8 mathrm{Li}_4(1/2)$$
          and also
          $$2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y = frac{pi^2}{12} - 8 mathrm{Li}_4(1/2) -7 log(2) zeta(3)+ frac{pi^2 log(2)^2}{3} - frac{ln(2)^4}{3} $$




          Finally, we see that eqref{2} can be written as
          $$frac{53}{360} pi^4 - 14 log(2) zeta(3) + frac{2}{3} [pi^2 -log(2)^2] log(2)^2 - 16 mathrm{Li}_4(1/2).$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
            $endgroup$
            – ben tenyson
            Jan 26 at 14:51








          • 4




            $begingroup$
            The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
            $endgroup$
            – ComplexYetTrivial
            Jan 26 at 15:04










          • $begingroup$
            The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
            $endgroup$
            – p4sch
            Jan 27 at 12:59












          • $begingroup$
            @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
            $endgroup$
            – ben tenyson
            Jan 27 at 13:26


















          3












          $begingroup$

          The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
          $$psi^{(m)}(x) = (-1)^{m+1} int_0^infty frac{t^m}{1-e^{-t}} e^{-xt} , mathrm{d} t.$$
          Using the monotone convergence theorem, we get
          $$label{1} tag{1}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} sum_{n=1}^infty frac{e^{-nt}}{2n-1} , mathrm{d}t. $$
          Since
          $$sum_{n=1}^infty e^{-(n-1/2)t} = frac{e^{t/2}}{e^t-1},$$
          we get by integration
          $$ sum_{n=1}^infty frac{e^{-nt}}{2n-1} = frac{e^{-t/2}}{2} int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s.$$
          But we have
          $$int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s = 2 int_{e^{t/2}}^infty frac{1}{x^2-1} , mathrm{d} x = 2 mathrm{arcoth}(e^{t/2}) $$
          and therefore
          $$sum_{n=1}^infty frac{e^{-nt}}{2n-1} = e^{-t/2} mathrm{arcoth}(e^{t/2}).$$
          Note that we have proven that $$mathrm{arcoth}(x) = sum_{n=1}^infty frac{x^{-(2n-1)}}{2n-1} quad text{for} quad |x| >1.$$




          Thus, we can rewrite eqref{1} by
          $$tag{2}label{2}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t.$$




          Changing variables we see that
          begin{align}
          - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t = - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x.
          end{align}
          Using $mathrm{artanh}(1/x) = mathrm{coth}(x)$ we can rewrite the integral also by
          $$ - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x= 8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y.$$




          With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.




          Using partial integration in the last line we get that
          begin{align}
          8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y &= left.-4 mathrm{artanh}(y)^2 ln(y)^2 right|_{y=0}^1 + 8 int_0mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y \
          &= 8 int_0^1mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y.
          end{align}
          Since $$mathrm{artanh}(y) = frac{1}{2} left( ln(y+1)-ln(1-y) right) $$
          we get that eqref{2} is equal to
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y - 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y + 2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y $$
          The remaining integrals are evaluated here, here and here. In fact, we have
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y = - zeta(4) = - frac{pi^4}{90} $$
          and
          $$- 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y = frac{3}{40} pi^4 - 7 log(2) zeta(3) + frac{pi^2 log(2)^2}{3}- frac{log(2)^4}{3} - 8 mathrm{Li}_4(1/2)$$
          and also
          $$2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y = frac{pi^2}{12} - 8 mathrm{Li}_4(1/2) -7 log(2) zeta(3)+ frac{pi^2 log(2)^2}{3} - frac{ln(2)^4}{3} $$




          Finally, we see that eqref{2} can be written as
          $$frac{53}{360} pi^4 - 14 log(2) zeta(3) + frac{2}{3} [pi^2 -log(2)^2] log(2)^2 - 16 mathrm{Li}_4(1/2).$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
            $endgroup$
            – ben tenyson
            Jan 26 at 14:51








          • 4




            $begingroup$
            The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
            $endgroup$
            – ComplexYetTrivial
            Jan 26 at 15:04










          • $begingroup$
            The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
            $endgroup$
            – p4sch
            Jan 27 at 12:59












          • $begingroup$
            @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
            $endgroup$
            – ben tenyson
            Jan 27 at 13:26
















          3












          3








          3





          $begingroup$

          The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
          $$psi^{(m)}(x) = (-1)^{m+1} int_0^infty frac{t^m}{1-e^{-t}} e^{-xt} , mathrm{d} t.$$
          Using the monotone convergence theorem, we get
          $$label{1} tag{1}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} sum_{n=1}^infty frac{e^{-nt}}{2n-1} , mathrm{d}t. $$
          Since
          $$sum_{n=1}^infty e^{-(n-1/2)t} = frac{e^{t/2}}{e^t-1},$$
          we get by integration
          $$ sum_{n=1}^infty frac{e^{-nt}}{2n-1} = frac{e^{-t/2}}{2} int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s.$$
          But we have
          $$int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s = 2 int_{e^{t/2}}^infty frac{1}{x^2-1} , mathrm{d} x = 2 mathrm{arcoth}(e^{t/2}) $$
          and therefore
          $$sum_{n=1}^infty frac{e^{-nt}}{2n-1} = e^{-t/2} mathrm{arcoth}(e^{t/2}).$$
          Note that we have proven that $$mathrm{arcoth}(x) = sum_{n=1}^infty frac{x^{-(2n-1)}}{2n-1} quad text{for} quad |x| >1.$$




          Thus, we can rewrite eqref{1} by
          $$tag{2}label{2}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t.$$




          Changing variables we see that
          begin{align}
          - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t = - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x.
          end{align}
          Using $mathrm{artanh}(1/x) = mathrm{coth}(x)$ we can rewrite the integral also by
          $$ - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x= 8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y.$$




          With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.




          Using partial integration in the last line we get that
          begin{align}
          8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y &= left.-4 mathrm{artanh}(y)^2 ln(y)^2 right|_{y=0}^1 + 8 int_0mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y \
          &= 8 int_0^1mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y.
          end{align}
          Since $$mathrm{artanh}(y) = frac{1}{2} left( ln(y+1)-ln(1-y) right) $$
          we get that eqref{2} is equal to
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y - 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y + 2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y $$
          The remaining integrals are evaluated here, here and here. In fact, we have
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y = - zeta(4) = - frac{pi^4}{90} $$
          and
          $$- 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y = frac{3}{40} pi^4 - 7 log(2) zeta(3) + frac{pi^2 log(2)^2}{3}- frac{log(2)^4}{3} - 8 mathrm{Li}_4(1/2)$$
          and also
          $$2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y = frac{pi^2}{12} - 8 mathrm{Li}_4(1/2) -7 log(2) zeta(3)+ frac{pi^2 log(2)^2}{3} - frac{ln(2)^4}{3} $$




          Finally, we see that eqref{2} can be written as
          $$frac{53}{360} pi^4 - 14 log(2) zeta(3) + frac{2}{3} [pi^2 -log(2)^2] log(2)^2 - 16 mathrm{Li}_4(1/2).$$







          share|cite|improve this answer











          $endgroup$



          The $n$-th derivate of the digamma function is known as "polygamma function of order $n$-th". Moreover, we have the integral representation
          $$psi^{(m)}(x) = (-1)^{m+1} int_0^infty frac{t^m}{1-e^{-t}} e^{-xt} , mathrm{d} t.$$
          Using the monotone convergence theorem, we get
          $$label{1} tag{1}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} sum_{n=1}^infty frac{e^{-nt}}{2n-1} , mathrm{d}t. $$
          Since
          $$sum_{n=1}^infty e^{-(n-1/2)t} = frac{e^{t/2}}{e^t-1},$$
          we get by integration
          $$ sum_{n=1}^infty frac{e^{-nt}}{2n-1} = frac{e^{-t/2}}{2} int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s.$$
          But we have
          $$int_t^infty frac{e^{s/2}}{e^s-1} , mathrm{d} s = 2 int_{e^{t/2}}^infty frac{1}{x^2-1} , mathrm{d} x = 2 mathrm{arcoth}(e^{t/2}) $$
          and therefore
          $$sum_{n=1}^infty frac{e^{-nt}}{2n-1} = e^{-t/2} mathrm{arcoth}(e^{t/2}).$$
          Note that we have proven that $$mathrm{arcoth}(x) = sum_{n=1}^infty frac{x^{-(2n-1)}}{2n-1} quad text{for} quad |x| >1.$$




          Thus, we can rewrite eqref{1} by
          $$tag{2}label{2}sum_{n=1}^infty frac{psi''(n)}{2n-1} = - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t.$$




          Changing variables we see that
          begin{align}
          - int_0^infty frac{t^2}{1-e^{-t}} e^{-t/2} mathrm{arcoth}(e^{t/2}) , mathrm{d} t = - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x.
          end{align}
          Using $mathrm{artanh}(1/x) = mathrm{coth}(x)$ we can rewrite the integral also by
          $$ - 8 int_1^infty frac{ln(x)^2}{x^2-1} mathrm{arcoth}(x) , mathrm{d} x= 8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y.$$




          With the help of ComplexYetTrivial's comment we can compute the last integral explicitly.




          Using partial integration in the last line we get that
          begin{align}
          8 int_0^1 mathrm{artanh}(y) frac{ln(y)^2}{y^2-1} , mathrm{d}y &= left.-4 mathrm{artanh}(y)^2 ln(y)^2 right|_{y=0}^1 + 8 int_0mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y \
          &= 8 int_0^1mathrm{artanh}(y)^2 frac{ln(y)}{y} , mathrm{d} y.
          end{align}
          Since $$mathrm{artanh}(y) = frac{1}{2} left( ln(y+1)-ln(1-y) right) $$
          we get that eqref{2} is equal to
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y - 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y + 2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y $$
          The remaining integrals are evaluated here, here and here. In fact, we have
          $$2 int_0^1 frac{ln(y+1)^2 ln(y)}{y} , mathrm{d} y = - zeta(4) = - frac{pi^4}{90} $$
          and
          $$- 4 int_0^1 frac{ln(y+1) ln(1-y) ln(y)}{y} , mathrm{d} y = frac{3}{40} pi^4 - 7 log(2) zeta(3) + frac{pi^2 log(2)^2}{3}- frac{log(2)^4}{3} - 8 mathrm{Li}_4(1/2)$$
          and also
          $$2 int_0^1 frac{ln(1-y)^2 ln(y)}{y} , mathrm{d} y = frac{pi^2}{12} - 8 mathrm{Li}_4(1/2) -7 log(2) zeta(3)+ frac{pi^2 log(2)^2}{3} - frac{ln(2)^4}{3} $$




          Finally, we see that eqref{2} can be written as
          $$frac{53}{360} pi^4 - 14 log(2) zeta(3) + frac{2}{3} [pi^2 -log(2)^2] log(2)^2 - 16 mathrm{Li}_4(1/2).$$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 27 at 15:12

























          answered Jan 26 at 12:29









          p4schp4sch

          5,460318




          5,460318












          • $begingroup$
            I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
            $endgroup$
            – ben tenyson
            Jan 26 at 14:51








          • 4




            $begingroup$
            The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
            $endgroup$
            – ComplexYetTrivial
            Jan 26 at 15:04










          • $begingroup$
            The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
            $endgroup$
            – p4sch
            Jan 27 at 12:59












          • $begingroup$
            @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
            $endgroup$
            – ben tenyson
            Jan 27 at 13:26




















          • $begingroup$
            I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
            $endgroup$
            – ben tenyson
            Jan 26 at 14:51








          • 4




            $begingroup$
            The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
            $endgroup$
            – ComplexYetTrivial
            Jan 26 at 15:04










          • $begingroup$
            The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
            $endgroup$
            – p4sch
            Jan 27 at 12:59












          • $begingroup$
            @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
            $endgroup$
            – ben tenyson
            Jan 27 at 13:26


















          $begingroup$
          I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
          $endgroup$
          – ben tenyson
          Jan 26 at 14:51






          $begingroup$
          I have found a link between this sum and this one $displaystyle sum_{n=1}^{infty}frac{psi^{0}(n+1/2)}{n^{3}}$ but I can't solve this one either could you give this one a try?
          $endgroup$
          – ben tenyson
          Jan 26 at 14:51






          4




          4




          $begingroup$
          The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
          $endgroup$
          – ComplexYetTrivial
          Jan 26 at 15:04




          $begingroup$
          The last integral can be computed using integration by parts and the definition of $operatorname{artanh}$ . The remaining integrals are evaluated here, here and here. The final result should be $$frac{53}{360} pi^4 + frac{2}{3}[pi^2 -ln(2)^2]ln(2)^2 - 14ln(2)zeta(3) -16 operatorname{Li}_4(1/2) , .$$
          $endgroup$
          – ComplexYetTrivial
          Jan 26 at 15:04












          $begingroup$
          The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
          $endgroup$
          – p4sch
          Jan 27 at 12:59






          $begingroup$
          The final result is correct. Also the numerical values for both the integral and the initial sum are identical with your result. :-) I have added your solution to my answer. (I have already tried a similar idea: Partial integration for $mathrm{arccoth}(x)$ and corresponding identity $mathrm{arccoth}(x) = ln((1+x)/(x-1))/2$, but the integral cannot be splitted, because three parts are divergent integrals, as we integrate from $x=1$ to $infty$.)
          $endgroup$
          – p4sch
          Jan 27 at 12:59














          $begingroup$
          @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
          $endgroup$
          – ben tenyson
          Jan 27 at 13:26






          $begingroup$
          @p4sch I encounted this sum while doing this double sum $sum_{n=1}^{infty}frac{1}{n^{3}}sum_{k=1}^{n}frac{1}{2k-1}$ .Changing the order of summation I get the sum in question but done without changing the order i get this $frac{1}{2}(frac{psi(n+1/2)}{n^{3}}-psi(frac{1}{2})zeta(3))$.Since this has been evaluated by ComplexyetTrivial sum of $frac{psi(n+1/2)}{n^{3}}$ can also be obtained.Special thanks to both of you.
          $endgroup$
          – ben tenyson
          Jan 27 at 13:26




















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