Let $\mathcal{H}=L^{2}[0,\infty)$. How can one easily show that $e^{-x}$ is a cyclic vector under the $C^{\star}$ subalgebra of operators on $\mathcal{L}(H)$ generated by all resolvents $(L-\lambda I)^{-1}$ of the selfadjoint operator $$ Lf=-\frac{d^{2}}{dx^{2}}f $$ whose domain $\mathcal{D}(L)$ consists of all twice absolutely continuous functions $f\in L^{2}[0,\infty)$ for which $f'' \in L^{2}[0,\infty)$ and $f(0)=0$.
What does the cyclic representation look like?
We can diagonalize $L$ (that is, make it a multiplication operator) with the help of the Fourier transform. More precisely, map $U:L^2(0,\infty)\to L^2(0,\infty)$, $Uf=\int f(x)\sin kx \, dx$; then $ULU^*$ is multiplication by $k^2$, and thus the cyclic vectors are exactly those for which $Uf\not=0$ almost everywhere. Clearly, $f(x)=e^{-x}$ has this property.