Computing $\int_0^T e^{itA}Be^{-itA} dt$ without an infinite series I'm hoping to compute the following integral: $\int_0^T e^{itA}Be^{-itA} dt$ where $iA, iB$ are traceless anti-Hermitian matrices (i.e. $\mathfrak{su}(n)$). I have found the following form for the integral: $\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{(iA)^n(iB)(iA)^m T^{n+m+1}}{n!m!(n+m+1)}$ by simply applying the Taylor series for the matrix exponential, however I can't find a simpler formula for the answer that doesn't involve an infinite series.
 A: I hope the following article can help: http://iopscience.iop.org/1063-7869/50/12/A02;jsessionid=EB2515B1F7F6CB626A8CBEF4537BBE05.c1 (Feynman disentangling of noncommuting operators and group representation theory, by V.S. Popov. A Russian version can be found here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=ufn&paperid=542&option_lang=eng). Namely, it seems, it follows from its equation (10.5) that $$\int_0^T e^{itA}Be^{-itA} dt=\ln{\left [e^{(iA+B)T}e^{-iAT}\right]}.$$
A: This is not an answer; rather, it is an extended comment on Terry's comments on Zurab's answer.
Here's an explicit counterexample to Popov's formula: Let
$$
iA = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad
B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} .
$$
Then the integrand equals
$$
\begin{pmatrix} e^t & 0 \\ 0 & e^{-t} \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} e^{-t} & 0 \\ 0 & e^t \end{pmatrix}
= \begin{pmatrix} 0 & -e^{2t} \\ e^{-2t} & 0 \end{pmatrix} \equiv F(t) .
$$
However, the matrix $C\equiv iA+B$ from Popov's formula equals
$$
C = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} ,
$$
so satisfies $C^2=0$, hence $e^{CT} = 1+CT$, and I think it's already very plausible that we cannot possibly have $e^{CT}e^{-iAT}$ equal to the exponential of the LHS. However, I also computed the LHS explicitly and found
$$
\exp \left( \int_0^T F(t)\, dt \right) = \begin{pmatrix} \cos\mu & (a/\mu)\sin\mu\\
-(\mu/a)\sin\mu & \cos\mu\end{pmatrix}, \quad a =-e^T\sinh T, \: \mu =\sqrt{\frac{\cosh 2T - 1}{2}} .
$$
Update: I think I've successfully read Popov's mind now: He's really claiming that (notation as above, $F$ is the integrand)
$$
Te^{\int_0^T F} = e^{(iA+B)T}e^{-iAT} \quad\quad (1)
$$
where (the first) $T$ is the time-ordering "operator" (see my comment above). More succinctly (and precisely), we define the LHS as the fundamental matrix of $Y'=YF$. Then (1) holds trivially (both sides solve the same IVP). While $\int F$ does make an appearance on the LHS, this is a purely notational sleight of hand and the formula really seems useless for the problem at hand.
A: For numerical purposes can you maybe treat it as a differential equation?
$$
u'(t) = e^{itA}Be^{-itA}, \qquad u(0) = 0
$$
A: I am not sure if you are ok with diagonalizing $A$ or if the size of the problem makes it unpractical. But if you can calculate eigenvalues and eigenvectors of $A$, and denote them as follows: $ A e_n = \lambda_n e_n$, then simply need to compute $e_m^H B e_n$ (where $e_m^H$ denotes the Hermitian transpose) for all $m$ and $n$:
$$ e_m^H \Big( \int_0^T e^{itA} B e^{-itA} dt \Big) e_n = e_m^H B e_n \  \Big(\int_0^T e^{it(\lambda_m - \lambda_n)} dt \Big)  $$
