Suppose $\{S_t\}_{t \geq 0}$ is an analytic semigroup on a Banach space $\mathcal{B}$ and $L$ is its generator. Also suppose that there are $M , a>0$ such that $\lVert S_t\rVert \leq Me^{-at}$ for all $t \geq 0$. For $\alpha >0$ define the negative fractional power of $L$ to be $$ (-L)^{-\alpha} := \frac{1}{\Gamma(\alpha)}\int_0^{\infty}t^{\alpha -1}S_t \,d t.$$
Now I have to show that, for $\alpha, \beta >0$,
$$(-L)^{-\alpha}(-L)^{-\beta}= (-L)^{-\alpha-\beta} .$$
From the definition I have $$ (-L)^{-\alpha}(-L)^{-\beta} = \frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^{\infty}\int_0^{\infty} t^{\alpha-1}u^{\beta-1} S_{t+u}\,d u\,d t $$ and this is where I'm stuck. I don't know whether I can show this just by manipulating the above integrals and using properties of the Gamma function or whether I need to use properties of the analytic semigroup. Thanks.
\begin{align} &\int_{0}^{\infty}t^{\alpha-1}S(t)\int_{0}^{\infty}u^{\beta-1}S(u)du dt \\ =&\int_{0}^{\infty}t^{\alpha-1}\int_{0}^{\infty}u^{\beta-1}S(u+t)dudt \\ =&\int_{0}^{\infty}t^{\alpha-1}\int_{t}^{\infty}(u-t)^{\beta-1}S(u)dudt \\ =&\int_{0}^{\infty} \int_{0}^{u}t^{\alpha-1}(u-t)^{\beta-1}dt S(u)du \end{align} The inner integral is a convolution integral, which can be found using Laplace transforms to be $$ t^{\alpha-1}\star t^{\beta-1}=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}t^{\alpha+\beta-1} $$ In fact, you can prove the above by replacing $S(u)$ with $e^{-us}$ in the first equations. And that gives you what you want: $(-L)^{\alpha}(-L)^{\beta}=(-L)^{\alpha+\beta}$.