Show that $\zeta(2,2) = \frac{3}{4} \zeta(4)$ by integrating $\int \frac{dx}{x} \wedge \frac{dx}{x} \wedge \frac{dx}{1-x} \wedge \frac{dx}{1-x}$ On Wikipedia there's an OK discussion of multiple zeta values (MZV).  We have an identity:
$$ \zeta(2,2) = \sum_{m > n > 0} \frac{1}{m^2 \, n^2}  = \frac{3}{4}  \sum_{n > 0} \frac{1}{ n^4}  = \frac{3}{4} \zeta(4) $$
This formula might already be on the Math.Stackexchange site (link?) Also notice that somehow the zeta function turns partly into an element of $\mathbb{Q}$.  I have already found an argument using one of the special cases of the shuffle formula.
$$ \zeta(2)^2 = 2 \, \zeta(2,2) + \zeta(4)  $$
Certainly, there still a search for even more $\zeta$-functions of this kind.  Both sides of this equation have integral formulas (due to Drinfiel'd):
\begin{eqnarray*} \zeta(2,2) &\stackrel{?}{=}& \int_{1 > x_1 > x_2 > x_3  > x_4 > 0}  
\frac{dx_1}{x_1} \wedge
\frac{dx_2}{1-x_2} \wedge
\frac{dx_3}{x_3} \wedge
\frac{dx_4}{1-x_4}  \tag{$*$}\\ \\
\zeta(4) &\stackrel{?}{=}& \int_{1 > x_1 > x_2 > x_3  > x_4 > 0}
\frac{dx_1}{x_1} \wedge \;\;
\frac{dx_2}{x_2} \;\;\wedge 
\frac{dx_3}{x_3} \wedge
\frac{dx_4}{1-x_4} \tag{$**$}
 \end{eqnarray*}
Even now, I'm not totally convinced the integrals on the right side, represent the infinite series on the left side.  But also, I wonder if there's a stronger notion of equivalence than just that they evaluate to the same number.  We have just shown that $\zeta(2,2)$ and $\zeta(4)$ are periods These notes of José Ignacio Burgos Gil and Javier Fresán indicate the following:

From the modern point of view, periods appear
  when comparing de Rham and Betti cohomology of algebraic varieties over
  number fields.

So, my other question is whether these two 4-forms are the same.  Possibly over $\mathbb{C} \backslash \{ 0,1\}$ or $\mathbb{P}^1 \backslash \{ 0,1, \infty\}$ :
\begin{eqnarray*} \omega_1 &=&   
\frac{dxt_1}{x_1} \wedge
\frac{dx_2}{1-x_2} \wedge
\frac{dx_3}{x_3} \wedge
\frac{dx_4}{1-x_4} \\ \\
\omega_2 &=& 
\frac{dx_1}{x_1} \wedge \;\;
\frac{dx_2}{x_2} \;\;\wedge 
\frac{dx_3}{x_3} \wedge
\frac{dx_4}{1-x_4}
 \end{eqnarray*}
And I would ask is there a sense in which $\omega_1 = \omega_2$?  Mostly I just want help with the integrals.
 A: If $f(m,n)=f(n,m)$, by symmetry
$$\sum_{m>n>0}f(m,n) = \frac{1}{2}\left[\Big(\sum_{n>0}f(n)\Big)^2-\sum_{n>0}f(n)^2\right] $$
hence
$$ \zeta(2,2) =\frac{\zeta(2)^2-\zeta(4)}{2}=\frac{\frac{5}{2}\zeta(4)-\zeta(4)}{2}=\frac{3}{4}\,\zeta(4)$$
since $\zeta(4)=\frac{2}{5}\zeta(2)^2$ can be proved by considering a suitable logarithmic integral and some combinatorial manipulations (see pages 7-8-9 of my notes), without even computing an explicit form for $\zeta(2)$ or $\zeta(4)$. Similar tricks can be implemented in a more general context through the residue theorem: see Flajolet and Salvy, Euler Sums and Contour Integral Representations.

($*$) and ($**$) have very simple proofs by substitutions:
$$ \int_{0<a<b<c<d<1}\frac{d\mu}{bcd(1-a)}=\int_{(0,1)^4}\frac{d\mu}{1-ABCD}=\sum_{n\geq 0}\int_{(0,1)^4}(ABCD)^n\,d\mu=\zeta(4) $$
$$\begin{eqnarray*}\int_{0<a<b<c<d<1}\frac{d\mu}{(1-a)b(1-c)d}&=&\int_{(0,1)^4}\frac{CD\,d\mu}{(1-ABCD)(1-CD)}\\&=&\int_{(0,1)^2}\frac{\text{Li}_2(CD)}{1-CD}\,d\mu\\&=&\sum_{n\geq 1}\frac{H_n^{(2)}}{(n+1)^2}=\zeta(2,2).\end{eqnarray*}$$
In general, Beuker-Calabi substitutions are pretty useful in showing equivalences between Euler sums, devising acceleration tricks, also estimating the irrationality measure of $\zeta(2),\zeta(3),\zeta(4)$.
