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.

$\endgroup$
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_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 $$\frac{53}{360} \pi^4 - 14 \log(2) \zeta(3) + \frac{2}{3} [\pi^2 -\log(2)^2] \log(2)^2 - 16 \mathrm{Li}_4(1/2).$$

| 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 '19 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 $$\frac{53}{360} \pi^4 + \frac{2}{3}[\pi^2 -\ln(2)^2]\ln(2)^2 - 14\ln(2)\zeta(3) -16 \operatorname{Li}_4(1/2) \, .$$ $\endgroup$ – 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.