TITLE: Integrating matrix exponential
QUESTION [6 upvotes]: I have a question about equation 6 in this paper.
Simplifying somewhat, the authors state the following
$$\int_0^{\infty} e^{-tL} dt = L^{-1}$$
$L$ here is a graph laplacian and therefore is a matrix. (Leave aside for the moment that $L$ is singular and thus make the above equation meaningless.)
To derive the equation, I first tried expanding $e^{-tL}$ into a series and integrating the individual terms in the hope that something clean comes out:
$$\int_0^{\infty} e^{-tL} dt = \sum_{i=0}^\infty \frac{(-L)^i}{i!}\int_0^{\infty} t^i dt$$
which lead nowhere for me.
So I tried the alternative tack of just treating $L$ as if it were scalar. In that case, the first equation is obviously true.
My question is, can I treat the matrix $L$ in the exponent as if it were a scalar and just integrate it?
(Bonus question, since $L$ is a graph laplacian and therefore singular, why are the authors inverting it when they should know better?)

REPLY [2 votes]: Following Will Jagy's suggestion, we start with $L = D_{uu} - W_{uu}$. In order to show $L$ is non-singular, it suffices to show that $D_{uu}^{-1} L= I- D_{uu}^{-1}W_{uu}$ is non-singular. 
Write $Q=D_{uu}^{-1}W_{uu}$. Then $Q$ has positive entries and the row sums are all strictly less than one. It's not hard to show that $Q^n\to 0$ exponentially fast, so that $(I-Q)^{-1}$ can be obtained directly as $\sum_{n=0}^\infty Q^n$. Done!