Is $\operatorname{tr}\left(\int A(t)\,dt\right) = \int \operatorname{tr}(A(t))\,dt$ true? Is it true that the trace of a matrix integration equals the integration of the trace of such matrix?
$$\operatorname{tr}\left(\int A(t)\,dt\right) = \int \operatorname{tr}(A(t))\,dt$$
 A: \begin{align}
\operatorname{tr} \int \begin{bmatrix} a_{11}(t) & \cdots & a_{1n}(t) \\ \vdots & & \vdots \\ a_{n1}(t) & \cdots & a_{nn}(t) \end{bmatrix} \, dt & = \operatorname{tr} \begin{bmatrix} \int a_{11}(t)\,dt & \cdots & \int a_{1n}(t) \,dt \\ \vdots & & \vdots \\ \int a_{n1}(t) \, dt & \cdots & \int a_{nn}(t)\,dt \end{bmatrix} \tag 1 \\[10pt]
& = \int a_{11}(t)\,dt + \cdots + \int a_{nn}(t)\,dt \tag 2 \\[10pt]
& = \int \Big(a_{11}(t) + \cdots + a_{nn}(t)\Big) \, dt \tag 3 \\[10pt]
& = \int\operatorname{tr} \begin{bmatrix} a_{11}(t) & \cdots & a_{1n}(t) \\ \vdots & & \vdots \\ a_{n1}(t) & \cdots & a_{nn}(t) \end{bmatrix} \, dt. \tag 4
\end{align}
Line $(1)$ makes an assumption about what an integral of a matrix-valued function means.
Line $(2)$ uses the definition of "trace".
Line$(3)$ uses linearity of the integral.
Line $(4)$ uses the definition of "trace".
A: If the integration is defined componentwise, it is certainly true -- the integral is the limit of a sum of matrices, and the trace of a sum is the sum of the traces. We can pass to the limit because trace is also continuous.
