Looking for a matrix A(t) I need your help, I'm looking for a contraexample, I need to give a matrix A(t), such that $$e^{\int_0^tA(s)ds}$$
is not a matrix solution for $x'=A(t)x$. I really don't have any clue what can it be. Thanks!
 A: Following is a concrete example where both the matrix exponential and the "integral"
( i.e. the fundamental solution of the corresponding matrix ODE ) is known:
Let $\sigma_3 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$, 
$\sigma_1 = \begin{bmatrix}0&1\\1&0\end{bmatrix}$ and consider 
$A(t) = t\sigma_3 + \sigma_1 = \begin{bmatrix}t & 1 \\ 1 & -t\end{bmatrix}$.
We have
$$\Lambda(t) \stackrel{def}{=} \int_0^t A(s) ds = \frac{t^2}{2}\sigma_3 + t\sigma_1$$
Notice $\;\Lambda(t)^2 = \lambda(t)^2 I_2\;$ where $\displaystyle\;\lambda(t) = t \sqrt{1 + \frac{t^2}{4}}\;$, the matrix exponential is given by
$$\begin{align}
{\large e^{\int_0^t A(s) ds} = e^{\Lambda(t)}}
&= \cosh\lambda(t) I_2 + \frac{\sinh\lambda(t)}{\lambda(t)}\Lambda(t)\\
&= \cosh\left(t\sqrt{1+\frac{t^2}{4}}\right) I_2 +
\frac{\sinh\left(t\sqrt{1+\frac{t^2}{4}}\right)}{\sqrt{1+\frac{t^2}{4}}}\left(\frac{t}{2}\sigma_3 + \sigma_1\right)
\end{align}
$$
The fundamental solution of the matrix ODE is the solution of following initial value problem.
$$X'(t) = A(t)X(t)\quad\text{ subject to }\quad X(0) = I_2\tag{*1}$$
Notice
$$X''(t) = A'(t) X(t) + A(t) X'(t) = ( A'(t) + A(t)^2 ) X(t) = (\sigma_3 + (1 + t^2)I_2) X(t)$$
Different entries of $X(t)$ decouple and they are solutions of ODE of the form
$$y''(t) = (\beta + 1 + t^2) y(t) \quad\text{ where }\quad \beta = \pm 1$$
The solutions of this sort of ODE are called Parabolic cylinder function. They can also be expressed
in terms of the more familiar confluent hypergeometric function. One can show that the fundamental solution 
of $(*1)$ is given by:
$$X(t) = e^{\frac{t^2}{2}}
\begin{bmatrix}
{}_1F_1(-\frac14;\frac12;-t^2) &
{}_1F_1(\frac14;\frac32;-t^2) t\\
{}_1F_1(\frac34;\frac32;-t^2) t &
{}_1F_1(\frac14;\frac12;-t^2)
\end{bmatrix}
$$
As one can see, even for something as simple as our $A(t)$ which is linear in $t$,
once $A(t)$ at different time fails to commute, the "integral" become very hard to
compute.
A: Hint: The trouble occurs because of non-commuting matrices.  Almost any example where $A(s) A(t) \ne A(t) A(s)$ will work.  Of course it would be good to take an example where the integral and the matrix exponential are easy to compute.
Try something upper or lower triangular.
