Can someone show the detailed procedures for proof:
\begin{equation*} \text{vec}\left(\int^T_0ds\,e^{-Ks}\Sigma\Sigma^\text{T}e^{-K^\text{T}s}\right) = \left(K\otimes I+I\otimes K\right)^{-1}\text{vec}\left(e^{-KT}\Sigma\Sigma^\text{T}e^{-K^\text{T}T}-\Sigma\Sigma^\text{T}\right). \end{equation*} Here $\text{vec}$ is vectorization operator for matrices.
It is amazing to be able to use vectorization technique.
Let $V$ be a finite dimensional complex vector space. Then functional calculus holds for linear operators $L(L(V))$ on the space of linear operators $L(V)$ on $V$, since $L(V)$ is finite dimensional. Consequently, if $X$ is an invertible element of $L(L(V))$ then
$$\int_0^T\mathrm{e}^{-sX}\,\mathrm{d}s=\frac{1-\mathrm{e}^{-TX}}{X}$$ by the functional calculus.
For any element $A$ of $L(V)$, write $A_{\bullet}$ for left multiplication by $A$ considered as a linear operator on $L(V)$, and likewise write $A^{\bullet}$ for right multiplication. Let $X=K_{\bullet}+(K^{\mathsf{T}})^{\bullet}$ in the expression above, assuming that this last expression is invertible. Then
$$\int_0^T\mathrm{e}^{-s(K_{\bullet}+(K^{\mathsf{T}})^{\bullet})}\,\mathrm{d}s=\frac{1-\mathrm{e}^{-T(K_{\bullet}+(K^{\mathsf{T}})^{\bullet})}}{K_{\bullet}+(K^{\mathsf{T}})^{\bullet}}$$
This equality is one between elements of $L(L(V))$. Applying both sides to the element $\Sigma\Sigma^{\mathsf{T}}$ of $L(V)$ gives $$\int_0^T\mathrm{e}^{-s(K_{\bullet}+(K^{\mathsf{T}})^{\bullet})}[\Sigma\Sigma^{\mathsf{T}}]\,\mathrm{d}s=\frac{1-\mathrm{e}^{-T(K_{\bullet}+(K^{\mathsf{T}})^{\bullet})}}{K_{\bullet}+(K^{\mathsf{T}})^{\bullet}}[\Sigma\Sigma^{\mathsf{T}}]$$ which simplifies to $$\int_0^T\mathrm{e}^{-sK}\Sigma\Sigma^{\mathsf{T}}\mathrm{e}^{-sK^{\mathsf{T}}}\,\mathrm{d}s=(K_{\bullet}+(K^{\mathsf{T}})^{\bullet})^{-1}[\Sigma\Sigma^{\mathsf{T}}-\mathrm{e}^{-sK}\Sigma\Sigma^{\mathsf{T}}\mathrm{e}^{-sK^{\mathsf{T}}}]$$ since left and right multiplication by elements of $L(V)$ commute as elements of $L(L(V))$.