Find: $\zeta \left( 3,1,1,1 \right)$

Question

While solving a summation, I came across this:

$$\zeta \left( 3,1,1,1 \right)=? $$

I'm new to multiple zeta values. That's why I couldn't find this. So my question is does a closed form exist for this. If yes, what is it? I have no idea to evaluate this. Please help.

Answer 1

It is known that every multiple zeta value can be expressed as a sum of MZVs of the same weight, where the arguments are 2's and 3's only. (Here weight means the sum of the arguments, so $ 1 + 1 + 1 + 3 = 6 $ in this case.) This is called the Hoffman basis, and this result is proven by Francis Brown in the paper "Mixed Tate motives over $\mathbb{Z}$".

The upshot is that you can express every MZV of weight 6 as a combination of $ \zeta(2,2,2) $ and $ \zeta(3,3) $. Then you can use that $ \zeta(2,2,2) = \frac{1}{7!} \pi^6 $ (see below), and the 'stuffle product' relation $ \zeta(3)^2 = \zeta(6) + 2\zeta(3,3) $. This allows you to get some combination of $ \pi^6 $ (or $ \zeta(6) $, or $ \zeta(2)^3 $ if you prefer) and $ \zeta(3) $ instead.

Specifically you find $$ \zeta(3,1,1,1) = -\frac{1}{2} \zeta(3)^2 + \frac{3}{4} \zeta(6) \, . $$

You can then use $ \zeta(6) = \frac{1}{945} \pi^6 $, if you want to write the result in terms of $ \pi $ instead.

You can use the tool EZ-Face to numerically confirm this, and this is perfectly sufficient to claim an identity at such a low weight. This command will find any linear dependencies between the inputs:

lindep([z(3)*z(3), z(6), z(3,1,1,1)])

The output is:

[2., -3., 4.]

This indicates that $ 2\zeta(3)^2 - 3\zeta(6) + 4 \zeta(3,1,1,1) = 0 $, which can be rearranged to the evaluation above.

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Edit: The result that $\zeta(2,2,2) = \frac{1}{7!} \pi^6 $, or more generally $$ \zeta(\underbrace{2, \ldots, 2}_{\text{\( k \) times}}) = \frac{1}{(2k+1)!} \pi^{2k} \, $$ can be found in Borwein, Bradley and Broadhurst's paper "Evaluations of $k$-fold Euler/Zagier sums". It follows by slightly simplifying equation (36).

You can find a proof for the specific case $\zeta(2,2,2)$ directly with the `stuffle product'. We have $$ \zeta(2)\zeta(2) \overset{\text{stuffle}}{=} \zeta(4) + 2 \zeta(2,2) $$ Multiply both sides by $ \zeta(2) $ and use $$ \zeta(2,2)\zeta(2) \overset{\text{stuffle}}{=} 3 \zeta(2,2,2) + \zeta(4,2) + \zeta(2,4) \, $$ to write $$ \zeta(2)^3 = \zeta(2)\zeta(4) + 6\zeta(2,2,2) + 2\zeta(4,2) + 2\zeta(2,4) \, . $$ Finally use $$ \zeta(4)\zeta(2) \overset{\text{stuffle}}{=} \zeta(6) + \zeta(4,2) + \zeta(2,4) $$ to get rid of the $ \zeta(4,2) $ and $ \zeta(2,4) $.

Now you have $$ \zeta(2)^3 = 3\zeta(2)\zeta(4) + 6 \zeta(2,2,2) - 2 \zeta(6) \, . $$ You can solve this for $ \zeta(2,2,2) $ and use Euler's evaluation of $ \zeta(\text{even}) $ to evaluate the result.

Answer 2

Applying the sum formula $$ \sum_{\sum a_i=n,\, a_i\ge 0}\zeta(a_1+2,a_2+1,...,a_r+1)=\zeta(n+r+1). $$ for $r=4$ and $n=1$ we obtain $$ \zeta(3,1,1,1)+\zeta(2,2,1,1)+\zeta(2,1,2,1)+\zeta(2,1,1,2) =\zeta(6). $$ I found in the literature that $\zeta(2,1,2,1)=\zeta(3,3)=\frac{\zeta(3)^2-\zeta(6)}{2}$, and similarly $$ \zeta(2,2,1,1)=\zeta(3)^2-\frac{4}{3}\zeta(6),\; \zeta(2,1,1,2)=\frac{25\zeta(6)-12\zeta(3)^2}{12}, $$ so that $$ \zeta(3,1,1,1)=\frac{3\zeta(6)-2\zeta(3)^2}{4}. $$