I was messing around in Wolfram Alpha when I stumbled on this closed form expression for the Hurwitz Zeta function:

$$ \zeta(3, 11/4) = 1/2 (56 \zeta(3) - 47360/9261 - 2 \pi^3). $$

How does WA know this? As far as I'm concerned there are no such closed forms in existence today. I am probably wrong, though.

My hunch is that it's related to Dirichlet characters and L-functions or some kind of polygamma function, but I can't prove it.

  • $\begingroup$ Wolfram Alpha only knows things that are in existence today (it's not creating new math on the fly—it's programmed using human knowledge). $\endgroup$ Commented 23 hours ago

3 Answers 3


Let's first rewrite $\zeta\left(3, \frac {11}4\right)$ as a polygamma function using : $$\tag{1}\psi^{(m)}(z) = (-1)^{m+1} m! \zeta(m+1,z)$$ We want $\displaystyle -\frac 12\psi^{(2)}\left( \frac {11}4\right)$ but from the recurrence relation $$\tag{2}\psi^{(m)}(z+1)= \psi^{(m)}(z) + \frac{(-1)^m m!}{z^{m+1}}$$ we have : \begin{align} \psi^{(2)}\left( \frac {11}4\right)&=\psi^{(2)}\left( \frac {7}4\right)+\frac{2}{\left( \frac {7}4\right)^3}\\ \tag{3}&=\psi^{(2)}\left( \frac {3}4\right)+\frac{2}{\left( \frac {3}4\right)^3}+\frac{2}{\left( \frac {7}4\right)^3}\\ \end{align} The first term is well known (see Kolbig ) : $$\tag{4}\psi^{(2n)}\left(\frac34\right) = 2^{2n-1}\left(\pi^{2n+1}|E_{2n}|-2(2n)!(2^{2n+1}-1)\zeta(2n+1)\right)$$ This gives (the second Euler number is $E_2=-1$) :


Combining all this gives us : $$\tag{6} \zeta\left(3,\frac{11}4\right)=-\frac 12\psi^{(2)}\left(\frac {11}4\right)=28\,\zeta(3)-\pi^{3}-\frac {23680}{9261}$$

  • $\begingroup$ Glad it helped @Klangen and excellent continuation! $\endgroup$ Commented 2 days ago
  • $\begingroup$ Excellent, once more ! $\endgroup$
    – Jean Marie
    Commented yesterday
  • $\begingroup$ Thanks @JeanMarie and use well the sunny days! :-) $\endgroup$ Commented yesterday

Tracing does not trace internals, but Mathematica yields

  Assuming[n \[Element] PositiveIntegers && q \[Element] Rationals, 
        HurwitzZeta[n, 11/4] // FunctionExpand // FullSimplify] 

$$\zeta \left(n,\frac{3}{4}\right)-\left(\frac{4}{3}\right)^n-\left(\frac{4}{7}\right)^n$$

      HurwitzZeta[7, 3/4] // FunctionExpand

$$\frac{2}{45} \left(182880 \zeta (7)-61 \pi ^7\right)$$

The general evalutations by Polygamma

    HurwitzZeta[3, q] // FunctionExpand

$$-(1/2) \text{PolyGamma[2, q]}=-\frac{\psi ^{(2)}(q)}{2}$$


We prove the equivalent statement $$\zeta(3,3/4)=\sum_{n=0}^\infty\frac1{(n+3/4)^3}=28\zeta(3)-\pi^3,$$ which we rewrite as $$\sum_{n=0}^\infty\frac1{(4n+3)^3}=\frac7{16}\zeta(3)-\frac{\pi^3}{64}.$$ Since $$\sum_{n=0}^\infty\frac1{(4n+1)^3}+\frac1{(4n+3)^3}=\frac78\zeta(3),$$ we need only prove that $$L(3,\chi_4)=\sum_{n=0}^\infty\frac1{(4n+1)^3}-\frac1{(4n+3)^3}=\frac{\pi^3}{32}.$$

To do this, we note that $\chi_4(n)=\sin(n\pi/2)$, and what we want to prove follows from the Fourier expansion of Bernoulli polynomials: $$B_{2r+1}(x)=(-1)^{r-1}2\cdot(2r+1)!(2\pi)^{-2r-1}\sum_{m=1}^\infty\frac{\sin(2\pi mx)}{m^{2r+1}}.$$ (See, for instance, Tenenbaum's Introduction to Analytic and Probabilistic Number Theory, exercise 2)


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