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.


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_0\mathrm{artanh}(y)^2 \frac{\ln(y)}{y} \, \mathrm{d} y \\ &= 8 \int_0^1\mathrm{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 $$\frac{53}{360} \pi^4 - 14 \log(2) \zeta(3) + \frac{2}{3} [\pi^2 -\log(2)^2] \log(2)^2 - 16 \mathrm{Li}_4(1/2).$$

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  • $\begingroup$ 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 '19 at 14:51
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    $\begingroup$ 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 $$\frac{53}{360} \pi^4 + \frac{2}{3}[\pi^2 -\ln(2)^2]\ln(2)^2 - 14\ln(2)\zeta(3) -16 \operatorname{Li}_4(1/2) \, .$$ $\endgroup$ – ComplexYetTrivial Jan 26 '19 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 '19 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 '19 at 13:26

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