Trying to understand some derivation in the paper: Deep Unsupervised Learning using Nonequilibrium Thermodynamics

I have recently been learning about diffusion models and trying to derive all the results in the paper by Sohl-Dickstein, et. al, "Deep Unsupervised Learning using Nonequilibrium Thermodynamics" (2015): https://arxiv.org/pdf/1503.03585.pdf In the Appendix B of this paper, they define the log-likelihood lower bound term as, $$K = \int{dx^{(0...T)}} q(x^{(0...T)}) \text{log} \Bigg[p(x^{(T)}) \prod_{t=1}^{T} \frac{p(x^{(t-1)}|x^{(t)})}{q(x^{(t)}|x^{(t-1)})} \Bigg]$$ . They further try to isolate the $p(x^{(T)})$ in the square bracket. The next step of the calculation should be, $$K = \int{dx^{(0...T)}} q(x^{(0...T)}) \text{log} \sum_{t=1}^{T}\Bigg[ \frac{p(x^{(t-1)}|x^{(t)})}{q(x^{(t)}|x^{(t-1)})} \Bigg] + \int{dx^{(0...T)}} q(x^{(0...T)}) \text{ log } p(x^{(T)})$$ The first integral above looks the same as the first term in the paper, I am trying to make the second integral above go to $$\int dx^{(T)} q(x^{(T)}) \text{ log } p(x^{(T)})$$ For this, I am using the fact that $q(x^{(0...T)})$ can be written as, $$q(x^{(0...T)}) = q(x^{(0)}) \prod_{t=1}^{T} q(x^{(t)}|x^{(t-1)})$$ Then I am writing the second integral above as, $$\int{dx^{(0...T)}} q(x^{(0...T)}) \text{ log } p(x^{(T)}) \\= \int dx^{(0)} q(x^{(0)}) \int dx^{(1)}q(x^{(1)}|x^{(0)}) \ldots \int dx^{(T-1)}q(x^{(T-1)}|x^{(T-2)}) \int dx^{(T)}q(x^{(T)}|x^{(T-1)}) \text{ log } p(x^{(T)}) \\ = \int dx^{(0)} q(x^{(0)}) \Bigg(\prod_{t=1}^{T-1} \int dx^{(t)}q(x^{(t)}|x^{(t-1)}) \Bigg) \int dx^{(T)}q(x^{(T)}|x^{(T-1)}) \text{ l…