Integral of inverse of Kronecker product Let $M$ be a real symmetric positive definite matrix of size $n\times n$. I would like to write the following matrix $L$ as a Kronecker product $A\otimes B$, for to be determined matrices $A$ and $B$. That is, I want $L = A\otimes B$. Is this possible?

$$L := \displaystyle\int_{0}^{\infty}\left(\left(M+\tau I\right)^{-1} \otimes \left(M+\tau I\right)^{-1}\right)\:\mathrm{d} \tau.$$

Attempt: from the properties of kronecker product, we know that 
$$ L = \displaystyle\int_{0}^{\infty}\left(\left(M+\tau I\right) \otimes \left(M+\tau I\right)\right)^{-1}\:\mathrm{d} \tau
= \displaystyle\int_{0}^{\infty} \left(M\otimes M + \tau(M\oplus M) + \tau^{2} I_{n^{2}}\right)^{-1}\:\mathrm{d} \tau.$$
Not sure how to go from here to $L = A \otimes B$ form. Another idea could be using Sherman-Morrison formula on $(M + \tau I)^{-1}$. If it helps, I am really interested to write $L^{-1}$ as a Kronecker product.
Special cases: If $M=I_{n}$, then $L = I_{n^{2}} = I_{n} \otimes I_{n}$. For diagonal $M = \mathrm{diag}(m_{1}, ..., m_{n})$ where $m_{i}>0$ for all $i=1,...,n$,  we have the $n^2 \times n^2$ matrix $L = \mathrm{diag}(L_{1}, ..., L_{n})$ with the $i$-th $n\times n$ diagonal matrix 
\begin{align}
L_{i} := \begin{cases}\displaystyle\int_{0}^{\infty}\frac{\mathrm{d}\tau}{(m_{i} + \tau)(m_{j}+\tau)} &= \displaystyle\frac{\log m_{i} - \log m_{j}}{m_{i} - m_{j}} & \text{for} \quad i\neq j,\\
\displaystyle\int_{0}^{\infty}\frac{\mathrm{d}\tau}{(m_{i} + \tau)^{2}} &= \displaystyle\frac{1}{m_{i}} & \text{for} \quad i=j.\end{cases}
\end{align} 
 A: Partial answer:
Motivated by the special cases noted in the question, let $M = Q\Lambda Q^{-1}$ be the spectral decomposition of $M \succ 0$. Since $\tau \geq 0$, we know $(M+\tau I) \succ 0$ (thus diagonalizable). Hence, $(M + \tau I)^{-1} = Q (\Lambda + \tau I)^{-1} Q^{-1}$ and
\begin{align}
L &= \displaystyle\int_{0}^{\infty}\left(\left(M+\tau I\right)^{-1} \otimes \left(M+\tau I\right)^{-1}\right)\:\mathrm{d} \tau\\
&= (Q \otimes Q) \left(\displaystyle\int_{0}^{\infty} (\Lambda + \tau I)^{-1} \otimes (\Lambda + \tau I)^{-1}\:\mathrm{d}\tau\right) (Q\otimes Q)^{-1},
\end{align}
where we repeatedly used the property that $AC \otimes BD = (A\otimes B)(C \otimes D)$. The matrix in the middle parenthesis is an $n^2\times n^2$ diagonal matrix whose elements are functions of $\lambda_{i} = {\rm{eig}}(M)$. Specifically,
\begin{align}
\left(\displaystyle\int_{0}^{\infty} (\Lambda + \tau I)^{-1} \otimes (\Lambda + \tau I)^{-1}\:\mathrm{d}\tau\right)_{ij} = \begin{cases}\displaystyle\frac{\log \lambda_{i} - \log \lambda_{j}}{\lambda_{i} - \lambda_{j}} & \text{for} \quad i\neq j,\\
\displaystyle\frac{1}{\lambda_{i}}& \text{for} \quad i=j.\end{cases}
\qquad\qquad(*)
\end{align}
Question: Is it possible to decompose (*) as a Kronecker product $\Lambda_{1}\otimes \Lambda_{2}$? If yes, we would indeed have $L = (Q \Lambda_{1} Q^{-1})\otimes (Q \Lambda_{2} Q^{-1})$ as desired.
It is interesting to note that we have the decomposition $L^{-1} = (Q \otimes Q) \left(\displaystyle\int_{0}^{\infty} (\Lambda + \tau I)^{-1} \otimes (\Lambda + \tau I)^{-1}\:\mathrm{d}\tau\right)^{-1} (Q\otimes Q)^{-1}$, where elements of the middle diagonal matrix, being element-wise inverses of (*), are logarithmic means between $\lambda_{i}$ and $\lambda_{j}$.
Edit: It is easy to show that $L^{-1}$ has the following integral representation:
$$L^{-1} = \displaystyle\int_{0}^{1} M^{\tau} \otimes M^{1-\tau} \mathrm{d}\tau = (Q\otimes Q) \left(\displaystyle\int_{0}^{1}\Lambda^{\tau} \otimes \Lambda^{1-\tau}\mathrm{d}\tau\right)(Q\otimes Q)^{-1}$$
