# How does Wolfram Alpha know this closed form?

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.

• 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). Commented 23 hours ago

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 $$E_2=-1$$) :

$$\tag{5}\psi^{(2)}\left(\frac34\right)=2\left(\pi^{3}-2(2)!(2^{3}-1)\zeta(3)\right)$$

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

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