= Solution
For the ordered <matrix exponential> product $\Phi(t)=e^{tA}e^{tB}$, differentiation gives
$$
\begin{aligned}
\Phi'(t)-(A+B)\Phi(t)
&=e^{tA}Be^{tB}-Be^{tA}e^{tB}\\
&=[e^{tA},B]e^{tB}.
\end{aligned}
$$
The <matrix commutator> here is $[X,Y]=XY-YX$. Put $E(t)=\Phi(t)-e^{t(A+B)}$. Then $E(0)=0$ and
$$
E'-(A+B)E=[e^{tA},B]e^{tB}.
$$
Multiplying on the left by $e^{-t(A+B)}$ gives an exact product derivative. Integration from zero to $t$, followed by multiplication by $e^{t(A+B)}$, proves the <exponential product defect identity>:
$$
\boxed{\Phi(t)=e^{t(A+B)}+
\int_0^t e^{(t-x)(A+B)}[e^{xA},B]e^{xB}\,dx.}
$$
No commutation of $A$ and $B$ was assumed. The ordered product is the usual <Lie-Trotter splitting>; the PDF calls this product the Beam-Warming splitting. If $[A,B]=0$, the commutator term vanishes and the product is exact. In general its small-step expansion has local error $t^2[A,B]/2+O(t^3)$.
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