I would like to know if the values of the derivative of the Riemann zeta function at negative odd integers are computed, i.e. $\zeta'(-n)$ when $n$ is odd. When I look at the page from Wolfram MathWorld or Wiki, I just found the value of $\zeta'(-1)$.

  • $\begingroup$ According to wikipedia the value $\zeta'(-1)$ is equal to $\frac{1}{12}-lnA$ where $A$ is the glaisher-kinkelin constant. $\endgroup$
    – Nicco
    Sep 26, 2015 at 10:28
  • $\begingroup$ @Nicco: You are right. I misread the wiki page. $\endgroup$
    – Paul
    Sep 26, 2015 at 10:58

2 Answers 2


In fact $\;\displaystyle\zeta(-1)=-\frac 1{12}\;$ while $\;\displaystyle\zeta'(-1)=\frac 1{12}-\ln A\;$ with $A$ the Glaisher–Kinkelin constant.

All this is given in Wikipedia including the general formula (for $\,n\in\mathbb{N}^*$) : $$\tag{1}\zeta'(-2n)=(-1)^n\frac{(2n)!}{2(2\pi)^{2n}}\;\zeta(2n+1)$$ Relations of this kind are usually obtained using the functional equation (with $s \leftrightarrow (1-s)$ here) : $$\tag{2}\zeta(1-s)=2(2\pi)^{-s}\Gamma(s)\,\zeta(s)\cos\left(\frac{\pi}2s\right)$$ the derivative gives us : \begin{align} \tag{3}-\zeta'(1-s)=\left(2(2\pi)^{-s}\Gamma(s)\,\zeta(s)\right)^{\,'}\,\cos\left(\frac{\pi}2s\right)+2(2\pi)^{-s}\Gamma(s)\,\zeta(s)\left(-\frac{\pi}2\sin\left(\frac{\pi}2s\right)\right)\\ \end{align} For $\;s=2n+1,\;n\in \mathbb{N}^*\,$ and since $\;\cos\left(\frac{\pi}2(2n+1)\right)=0,\;\sin\left(\frac{\pi}2(2n+1)\right)=(-1)^n\frac{\pi}2\,$ we get : \begin{align} \zeta'(-2n)=\pi\,(2\pi)^{-(2n+1)}\Gamma(2n+1)\,\zeta(2n+1)(-1)^n\\ \end{align} i.e. the wished relation $(1)$.

Now you were interested by the case $s=2n,\;n\in \mathbb{N}^*$ (with $1-2n\,$ an odd and negative integer) but in this case $(3)$ is not so simple (we will use $\Gamma'(s)=\Gamma(s)\psi(s)$ with $\psi$ the digamma function) : \begin{align} -\zeta'(1-2n)&=\left.\left(2(2\pi)^{-s}\Gamma(s)\,\zeta(s)\right)^{\,'}\right|_{s=2n}\,\cos\left(\pi\,n\right)\\ \tag{4}\zeta'(1-2n)&=(-1)^{n+1}\frac{2\,\Gamma(2n)}{(2\pi)^{2n}}\left((-\log(2\pi)+\psi(2n))\zeta(2n)+\zeta'(2n)\right)\\ \end{align}

From this we see (for example) that $\zeta'(-1)$ may be written in function of $\zeta(2)$ and $\zeta'(2)$ and may thus justify the alternative formula for the GK constant (using $\psi(2)=1-\gamma$ the Euler constant) : $$-\zeta'(2)=\zeta(2)\left(12\log(A)-\gamma-\log(2\pi)\right)$$ but all this shows only that there are no simple expressions for $\zeta'($negative odd$)$!

Other relations (after $(4)$ in the link) may be obtained using the logarithmic derivative of $(2)$, there are too some curious relations for $\zeta(3)$ and $\zeta(1/2)$ at MO but this shouldn't help here.

Hoping this clarified things any way,

  • $\begingroup$ Excellent post. The lack of a simple expression for these values only makes them more interesting. Fascinating numbers. For example: every other negative odd integer value of zeta (-1, -5, -9 . . . ) is a negative fraction with a denominator that is divisible by 12, and the denominator 12 itself seems to repeat in a semi-periodic fashion as the absolute values of the results grow larger. $\endgroup$
    – emery
    Apr 6, 2019 at 0:47
  • $\begingroup$ Thanks @emery! Concerning $\;\zeta(-1-4n)\,$ we have $\;\zeta(-1-4n)=-\dfrac{B_{4n+2}}{4n+2}\,$ linking your observations to the always fascinating properties of the Bernoulli numbers. Excellent discoveries! $\endgroup$ Apr 6, 2019 at 10:20

For where $m∈ℤ^+$,


$$ζ′(1-2m) = \frac{1}{2m}ψ(2m)B_{2m}-(2m-1)!∫_1^{∞}\frac{z^{-2m}}{e^{z}-1}dz-\frac{(2m-2)!}{2} $$ $$-(2m-1)!∑_{t=0,t\neq m}^{∞}\frac{B_{2t}}{(2t)!}\frac{1}{2t-2m}$$

where $ψ$ is the digamma function and $B_{n}$ is the $n$-th Bernoulli number.


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