# $\psi(2,1/6),\psi(4,1/6)$ in terms of zeta and pi only and another closed form for zeta

Let $$\psi(n,x)$$ denote the polygamma function.

In this answer Lucia gave linear relations for $$\psi(m,1/3),\psi(m,1/6),\zeta(m+1)$$.

The computer managed to find closed form for $$\psi(2,1/6)$$ and $$\psi(4,1/6)$$ in terms of $$\zeta, \pi$$ and $$\sqrt{3}$$. As a byproduct this give another closed form for $$\zeta(3),\zeta(5)$$.

Working with precision 1000 decimal digits we have:

$$\psi(2,1/6)= -2/3 \sqrt{3}(6\pi^3 + 91\sqrt{3}\zeta(3))$$ $$\psi(4,1/6)=-176 \sqrt{3}\pi^5 - 90024\zeta(5)$$ $$\zeta(3)=-1/546\sqrt{3}(12 \pi^3 + \sqrt{3}\psi(2, 1/6))$$

$$\zeta(5)= -2/1023 \sqrt{3} \pi^5 - 1/90024 \psi(4, 1/6)$$

Are these experimental results correct? Do they generalize for $$\psi(m,1/6)$$ for other $$m$$?

Added Looks like for all $$m$$, $$\pi,\psi(m,1/6),\psi(m,1/3)$$ are algebraically dependent and if we add $$\sqrt{3}$$ they are linearly dependent. Proving this and Lucia's answer will answer the question.

 p_2_16= -2/3*sqrt(3)*(6*pi^3 + 91*sqrt(3)*zeta(3)) # psi(2,1/6)
p_4_16=-176*sqrt(3)*pi^5 - 90024*zeta(5) # psi(4,1/6)

zeta3=-1/546*sqrt(3)*(12*pi^3 + sqrt(3)*psi(2, 1/6))
zeta5= -2/1023*sqrt(3)*pi^5 - 1/90024*psi(4, 1/6)

print 'psi(2,1/6)',(p_2_16 - psi(2,1/6))
print 'psi(4,1/6)',p_4_16 - psi(4,1/6)
print 'zeta(3)',zeta3 - zeta(3)
print 'zeta(5)',zeta5 - zeta(5)

#linear depend.
m= 2 ; x0=psi(2, 1/6); x1=psi(2, 1/3); x3=sqrt(3)*pi^3 ;f= 9*x0 - 63*x1 + 8*x3
m= 3 ; x0=psi(3, 1/6); x1=psi(3, 1/3); x2=pi^4 ;f= 3*x0 - 51*x1 + 16*x2
m= 4 ; x0=psi(4, 1/6); x1=psi(4, 1/3); x3=sqrt(3)*pi^5 ;f= 3*x0 - 93*x1 + 32*x3
m= 5 ; x0=psi(5, 1/6); x1=psi(5, 1/3); x2=pi^6 ;f= 9*x0 - 585*x1 + 832*x2
m= 6 ; x0=psi(6, 1/6); x1=psi(6, 1/3); x3=sqrt(3)*pi^7 ;f= 3*x0 - 381*x1 + 896*x3
m= 7 ; x0=psi(7, 1/6); x1=psi(7, 1/3); x2=pi^8 ;f= -3*x0 + 771*x1 - 10496*x2
m= 8 ; x0=psi(8, 1/6); x1=psi(8, 1/3); x3=sqrt(3)*pi^9 ;f= 27*x0 - 13797*x1 + 414208*x3
m= 9 ; x0=psi(9, 1/6); x1=psi(9, 1/3); x2=pi^10 ;f= 3*x0 - 3075*x1 + 687104*x2
m= 10 ; x0=psi(10, 1/6); x1=psi(10, 1/3); x3=sqrt(3)*pi^11 ;f= -3*x0 + 6141*x1 - 3782656*x3
m= 11 ; x0=psi(11, 1/6); x1=psi(11, 1/3); x2=pi^12 ;f= 9*x0 - 36873*x1 + 206614528*x2

• Where the term sqrt(3) came from? Just shooting in the dark, but could this be \psi doubling formula? – joro Oct 11 at 15:02

These relations are well-understood and can be proved rigorously. Recall that the polygamma function $$\psi(n,x)$$ can be expressed in terms of the Hurwitz zeta function (see here): $$\psi(m,x)=(-1)^{m+1} m! \cdot \zeta(m+1,x) \quad \textrm{where} \quad \zeta(m+1,x)=\sum_{n=0}^{\infty} \frac{1}{(x+n)^{m+1}}.$$ In particular for $$x=a/N$$ with $$0 and $$(a,N)=1$$, we get $$\psi(m,a/N)=(-1)^{m+1} m! \cdot \zeta(m+1,a/N)=(-N)^{m+1} m! \sum_{\substack{n \geq 1 \\ n \equiv a (N)}} \frac{1}{n^{m+1}}.$$ These series is a particular case of the Dirichlet series $$\sum_{n \geq 1} \theta(n)/n^s$$ where $$\theta : \mathbb{Z} \to \mathbb{C}$$ is $$N$$-periodic. In our case $$\theta$$ is the characteristic function $$\theta_a$$ of $$\bar{a} \in \mathbb{Z}/N\mathbb{Z}$$. Now you can decompose $$\theta_a$$ as a linear combination of Dirichlet characters modulo $$N$$, which gives $$\psi(m,a/N)=\frac{(-N)^{m+1} m!}{\varphi(N)} \sum_{\chi \textrm{ mod } N} \overline{\chi}(a) L(\chi,m+1).$$ The artihmetic nature of the special value $$L(\chi,m+1)$$ is very different according to the parity of $$\chi$$. If $$\chi$$ and $$m+1$$ have the same parity, that is $$\chi(-1)=(-1)^{m+1}$$, then $$L(\chi,m+1)$$ is a critical value in the sense of Deligne (the $$\Gamma$$-factor of $$L(\chi,s)$$ has no pole at $$s=-m$$) and $$L(\chi,m+1)$$ is an algebraic multiple of $$\pi^{m+1}$$. The expression is explicit in terms of generalized Bernoulli numbers. The factor $$\sqrt{3}$$ that you found out is essentially the Gauss sum of the non-trivial character modulo $$3$$.
On the other hand, if $$\chi$$ and $$m+1$$ have opposite parity, then $$L(\chi,m+1)$$ is non-critical, the simplest example being $$\zeta(3)$$. In this case no expression in terms of usual special functions is known, but there does exist a nice theory about these $$L$$-values: they are related to the Borel regulator on the $$K$$-group $$K_{2m+1}(F_\chi)$$, where $$F_\chi \subset \mathbb{Q}(\zeta_N)$$ is the abelian number field cut out by the kernel of $$\chi$$. See for example this article by Zagier.
Of course, it may happen that suitable linear combinations of $$\theta_a$$'s are even or odd, which explains the formulas involving only a power of $$\pi$$.
• Thanks. Is it known for which $x$ we have zeta(3) = f(psi(2,x),simpler terms)? – joro Oct 11 at 14:20
• @joro If you mean $\zeta(3)$ expressed in terms of a single $\psi(2,a/N)$ and other simpler terms, then I guess a necessary condition is that the group $(\mathbb{Z}/N\mathbb{Z})^\times/\pm 1$ is trivial, so $N=2,3,4,6$. – François Brunault Oct 11 at 14:44
• @joro Yes, this generalizes to zeta at any odd integer. The term $\pi^3/\sqrt{3}$ in $\psi(2,1/3)$ and $\psi(2,1/6)$ comes from the non-trivial character $\chi$ mod 3 (or 6). It is known in general that $L(\chi,1-m) \in \mathbb{Q}(\chi) = \mathbb{Q}$ here. Now the functional equation gives $L(\chi,m) = (\textrm{some factor}) (2\pi i)^m/G(\bar{\chi}) L(\chi,1-m)$ where $G$ is the Gauss sum. This explains the $\sqrt{3}$. – François Brunault Oct 11 at 15:40
• @joro In a more elementary way, you can avoid the functional equation and prove directly these identities by considering the Fourier expansion of the (1-periodic version of the) Bernoulli polynomials, and use the fact that the discrete Fourier transform of $\chi$ as a function on $\mathbb{Z}/N\mathbb{Z}$ is $G(\chi) \bar{\chi}$ if $\chi$ is primitive. In this way you get $L(\chi,m) \sim (2\pi i)^m/G(\bar{\chi})$ if $\chi(-1)=(-1)^m$. – François Brunault Oct 11 at 16:00