# 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.

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 $$\Omega_n(z)_{21} = -\Omega_n(z)_{12}$$.
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?