Infinite sum of iterated integrals of matrix products Edit: Discussion moved to Mathoverflow at https://mathoverflow.net/questions/395085/infinite-sum-of-iterated-integrals-of-matrix-products
The problem:
Let
$$N(z) = \begin{pmatrix} 0 & \frac{1}{z} \\ \frac{1}{2(z-1)^2} & 0 \end{pmatrix} \;.$$
How do I find a closed form expression for the following infinite sum over permutations?
$$\Omega(z) =\sum_{n=1}^{\infty} \sum_{\sigma \in S_n} \omega_\sigma \; M_\sigma(z) \; .$$
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 only obvious pattern for the $n$th term of $\Omega(z)$ that I have noticed is that the diagonal entries are equal up to a minus sign for even $n$.
Edit: The pattern above was proved nicely in the comment of Joshua P. Swanson.
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.
 A: Here's a proof of your diagonal entry observation that's too long for a comment. Let $P = \begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then $N(z_1)N(z_2) = PN(z_2)N(z_1)P$. When $\sigma \in S_n$ for $n$ even, it follows that $$PN(z_{\sigma(1)}) \cdots N(z_{\sigma(n)})P = N(z_{\sigma(n)}) \cdots N(z_{\sigma(1)}) = N(z_{\overline{\sigma}(1)}) \cdots N(z_{\overline{\sigma}(n)})$$
where $\overline{\sigma}(i) := \sigma(n-i+1)$ is the reversal of $\sigma$. Clearly $d_{\overline{\sigma}} = n-1-d_{\sigma}$, so $\omega_{\overline{\sigma}} = -\omega_\sigma$. Letting $\Omega_n(z) := \sum_{\sigma \in S_n} \omega_\sigma M_\sigma(z)$, it follows that
$$P\Omega_n(z)P = -\Omega_n(z)$$
Conjugating by $P$ rotates the matrix entries 180 degrees, so it swaps the main diagonal and off diagonal entries, respectively. Hence $\Omega_n(z)_{11} = -\Omega_n(z)_{22}$ and $\Omega_n(z)_{21} = -\Omega_n(z)_{12}$.
This overall argument suggests your non-commutative determinants still somehow have a good amount of symmetry. I have little expertise in statistical mechanics, integrable systems, and the like. You may get a better answer at MathOverflow.
It would probably be wise to include a little motivation, at least at the end. Otherwise who's to say you didn't mess up some factor in $\omega_\sigma$, say, which would probably make the question nonsense?
