1. Question

Could anyone recommend a useful method for approaching the following series?

$$\sum_{k=2}^{\infty}\zeta'(k)x^{k-1}$$ Where $\zeta(z)$ is the Riemann Zeta function.

I've seen that there are many series involving this function, but I haven't found any that involve its derivative, the analogue to this series without the derivative is


which has a closed form solution, I was curious to know if there was a closed form for the version with the derivative too

Any suggestions are welcome


I arrived at this formulation: $$\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n-1}=-\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{\ln\left(n\right)}{k^{n}}x^{n-1}=-\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{\ln\left(n\right)}{k^{n}}x^{n-1}=\frac{1}{x}\sum_{k=1}^{\infty}\text{Li}_{0}^{\left(1,0\right)}\left(\frac{x}{k}\right)$$


$$\text{Li}_{0}^{\left(1,0\right)}\left(z\right):=\left.\frac{\partial}{\partial \nu}\text{Li}_{\nu}\left(z\right)\right|_{\nu=0}=-\sum_{k=1}^{\infty}\ln\left(k\right)z^{k}$$


$$\sum_{n=2}^{\infty}\zeta'\left(n\right)x^{n-1}=\frac{1}{x}\frac{\partial}{\partial \nu}\left.\left[\sum_{k=1}^{\infty}\text{Li}_{\nu}\left(\frac{x}{k}\right)\right]\right|_{\nu=0}$$

  • 1
    $\begingroup$ I tried to look into this and a similar series about a year ago out of curiosity, and I don't believe there exists much of a closed form. However, I maybe remember finding some alternate expressions. I'll have a look through my stack of discarded work to see if I can find anything later... $\endgroup$ Nov 13, 2023 at 22:40
  • $\begingroup$ @C-RAM Thank you very much! $\endgroup$ Nov 13, 2023 at 22:43
  • 1
    $\begingroup$ Maybe when $x=-1$, you get the Lüroth/Allani-Grinstead constant $\endgroup$ Nov 13, 2023 at 22:46
  • $\begingroup$ Well, I have found a portion of my old work, and it appears that what I had found was a somewhat non-trivial alternate expression for the related series $\sum_{k=2}^\infty\zeta'(2k) x^{2k}$ involving an integral of the log-$\Gamma$ function. There should also be a somewhat trivial expression for $\sum_{k=2}^\infty\zeta'(k) x^k$ as a pole expansion, which allows for analytic continuation of the series. I'm not 100% sure on the details, so do let me know if any of those things would interest you either as a comment or as an answer, so I can try hash out those details and type something up. $\endgroup$ Nov 15, 2023 at 8:44
  • $\begingroup$ @C-RAM Yes, I'm very interested and it would be very good as an answer (I was hypothesizing that the function could depend on some integral of the Gamma function but I wasn't sure) $\endgroup$ Nov 15, 2023 at 10:35

1 Answer 1


This is not a full answer as it does not really provide a "closed form" representation of the series, though I highly suspect no such closed form exists. Rather, here is short discussion on some different forms of this and a related series that could be of some interest to the asker.

Firstly, we have the following easy to prove identity \begin{equation} \boxed{\sum_{k=2}^\infty \zeta'(k)x^k=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}} \end{equation} for $|x|<2$ which provides the analytic continuation of the first sum to the whole complex plane except for integers $x\geq 2$.

To see this, write

\begin{equation} \begin{split} \sum_{k=2}^\infty \zeta'(k)x^k&=\sum_{k=2}^\infty\left[\sum_{n=2}^\infty\frac{\log(n)}{n^k}\right]x^k\\ &=\sum_{n=2}^\infty\log(n)\sum_{k=2}^\infty\left(\frac{x}{n}\right)^k\\ &=\sum_{n=2}^\infty\frac{\log(n)x^2}{n(n-x)}\\ \end{split} \end{equation}

The related series representing the even part of the series discussed above admits the following representation \begin{equation} \boxed{\sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx} \end{equation} for $|y|<2$ (for $y\in\{-1,0,1\}$, the right-hand expression can be interpreted as a limit). Note that the integral converges for any complex $y$, so this identity again provides an analytic continuation of the sum to the whole complex plane except for integers $y$ with $|y|\geq 2$.

We begin with the following well known Fourier series

\begin{equation} \begin{split} \log\Gamma(x)&=\frac12\log(2\pi)+\sum_{k=1}^\infty \frac{1}{2k}\cos(2k\pi x)+\sum_{k=1}^\infty \frac{1}{\pi k}[\gamma+\log(2k\pi)]\sin(2k\pi x)\\ B_{2n-1}(x)&=2(2n-1)!(-1)^{2n-1}\sum_{k=1}^\infty \frac{\sin(2k\pi x)}{(2k\pi)^{2n-1}}\\ \end{split} \end{equation} for $0<x<1$ and $n\geq 1$, where $B_m(x)$ is the $m$-th Bernoulli polynomial. Now, by using the above series and appealing to the orthogonality relationships between sin and cos, we may show that. \begin{equation} \int_0^1 B_{2n-1}(x)\log\Gamma(x)dx=2(-1)^n\frac{(2n-1)!}{(2\pi)^{2n}}[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)] \end{equation} for $n\geq 1$. Finally, using the generating function identity

$$\frac{t \sinh((x-1/2)t)}{2\sinh(t/2)}=\sum_{n=1}^\infty B_{2n-1}(x)\frac{t^{2n-1}}{(2n-1)!}$$

we have that \begin{equation} \begin{split} \int_0^1 \frac{t\sinh((x-1/2)t)}{2\sinh(t/2)}\log\Gamma(x)dx&=2\sum_{n=1}^\infty (-1)^n[(\gamma+\log(2\pi))\zeta(2n)-\zeta'(2n)]\frac{t^{2n-1}}{(2\pi)^{2n}}\\ &=[\gamma+\log(2\pi)]\left(\frac1{t}+\frac{i}{2}\cot\left(\frac{t}{2i}\right)\right)-\frac2{t}\sum_{n=1}^\infty\zeta'(2n)\left(\frac{t}{2\pi i}\right)^{2n}\\ \end{split} \end{equation} Performing the change of variable $t\mapsto 2\pi iy$, we have that \begin{equation} \int_0^1 \frac{\pi i y \sin(\pi y(2x-1))}{\sin(\pi y)}\log\Gamma(x)dx=[\gamma+\log(2\pi)]\left(-\frac{i}{2\pi y}+\frac{i}{2}\cot(\pi y)\right)+\frac{i}{\pi y}\sum_{n=1}^\infty\zeta'(2n)y^{2n} \end{equation} taking the imaginary part of both sides, and rearranging, we have that \begin{equation} \sum_{n=1}^\infty\zeta'(2n)y^{2n}\!=\!\frac1{2}[\gamma+\log(2\pi)](1-\pi y\cot(\pi y))+\pi^2y^2\csc(\pi y)\!\!\int_0^1\!\!\!\sin(\pi y(2x-1))\log\Gamma(x)dx \end{equation} as desired.

  • $\begingroup$ An observation in case you also want to work with the series for odd numbers: $$\frac{1-\pi y\cot(\pi y)}{2}=\frac{\psi\left(1+y\right)-\psi\left(1-y\right)}{2}y$$ which recalls the structure of an imaginary part (ditto regarding the sine which is the imaginary part of the exponential), I'm not sure but perhaps in the other formula a $\dfrac{\psi\left(1+y\right)+\psi\left(1-y\right)}{2}y$ and a $\cos(\pi y(2x-1))$ could appear $\endgroup$ Nov 23, 2023 at 10:36
  • 1
    $\begingroup$ @MathAttack I did very briefly look into trying to rework this for the series with odd $\zeta'$ values, and maybe some other method would work for it, but using this method, an issue arises: The even Bernoulli polynomials $B_{2n}(x)$ have a fourier series only involving cos, and the analogous sin series has no closed form in general (that I'm aware of), so we don't have a way to extract the $\frac{\log(k)}k$ terms from the Fourier series of $\log\Gamma(x)$. I have one or two ideas that might work, but nothing concrete. I may look into them if I find time. $\endgroup$ Nov 23, 2023 at 11:22
  • $\begingroup$ The OP has found the solution and posted it on his Instagram Page. Informing you on behalf of Math Attack. $\endgroup$ Feb 10 at 14:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .