Originally asked over at Stackexchange (https://math.stackexchange.com/questions/4169812/infinite-sum-of-iterated-integrals-of-matrix-products), but this forum was deemed more appropriate.

The problem:

Let

$$N(z) = \begin{pmatrix} 0 & \frac{1}{z} \\ \frac{1}{2(z-1)^2} & 0 \end{pmatrix} \;.$$

I write this post to ask: How does one obtain a closed form for the following infinite sum over permutation?

$$\Omega(z) =\sum_{n=1}^{\infty} \sum_{\sigma \in S_n} \omega_\sigma \; M_\sigma(z) \; .$$

Here $M_\sigma(z)$ is an $n$-fold iterated integral of matrix products (the lower integration bounds vanish):

$$M_\sigma(z) = \int^z dz_1 \int^{z_1} dz_2 \cdots \int^{z_{n-1}} dz_n \; N(z_{\sigma(1)}) \cdots N(z_{\sigma(n)})$$

The coefficients $\omega_\sigma$ are:

$$\omega_\sigma = \left(-\frac{i}{\sqrt{2}} \right)^n \frac{(-1)^{d_\sigma}}{n \; \binom{n-1}{d_\sigma}} \;.$$

The $d_\sigma$ above counts the number of descents of a permutation $\sigma$ acting on $\{1,2,\ldots,n\}$.

A permutation has a descent at position $i$ if $\sigma(i)>\sigma(i+1)$. E.g if $\{\sigma(1),\sigma(2),\sigma(3)\} = \{3,2,1\}$, then $d_\sigma = 2$ because there are descents at $i=1,2$.

What I know so far

I have very good reason to suspect that all four entries of $\Omega(z)$ will resum to some combination of elliptic integrals $K(z)$ and $E(z)$, so perhaps any identities related to these functions could be a promising line of attack.

The even terms in $\Omega(z)$ (i.e for $n=2m$) have the form $\quad \begin{pmatrix} f_n & 0 \\ 0 & -f_n \end{pmatrix}$

Motivation:

This is an example of computing a Magnus transformation, using equation (18) in https://iopscience.iop.org/article/10.1088/2399-6528/aab291/pdf

The matrix $\Omega$ can be used to solve a 1st order, linear matrix DEQ (see equations (1)-(3) in the article). Such DEQs appear in various contexts in quantum mechanics.