Free particle Schrodinger equation: propagator

I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $$U(t)$$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $$E$$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$ where $$\lvert E, \alpha \rangle$$ are the orthonormal eigenkets for the $$E$$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$

I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $$\lvert E\rangle = \lvert p\rangle$$ and $$p = \pm\sqrt{2mE}$$, he chooses the propagator as a function of the momentum eigenvalues $$p$$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $$-p$$, and the bounds switch from $$0$$ to $$\infty$$ such that: $$U(t) = \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $$\lvert p\rangle = \lvert E, + \rangle$$, $$\lvert -p\rangle = \lvert E, - \rangle$$, $$E = p^2/2m$$, and $$dp = \pm mdE/\sqrt{2mE}$$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}.$$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $$m/\sqrt{2mE}$$ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues?

Your first expression should have a additional factor in the integrand called a density of states (DOS) and is usually denoted $\rho(E)$, i.e., $$U(t)=\sum\int\rho(E)dE|E,\alpha\rangle\langle E,\alpha|e^{-iEt/\hbar}.$$ The density of states is the number of states in the energy range $(E,E+dE)$. Look at the parabolic dispersion section of the density of states on wiki.