# Floating swap payoff with rate determined on current instead of previous date

I am attempting to determine the payoffs a modified swap, in which the floating payments at a time $$T_k$$ are made on the current date (i.e. $$L(T_k,T_{k+1})\equiv L_{k+1}(T_k)$$) rather than at the previous date $$T_{k-1}$$ (i.e. the usual $$L(T_{k-1},T_k)\equiv L_k(T_{k-1})$$). I want to prove that that the total time-$$t$$ payoff (with $$\delta_k=T_k-T_{k-1}$$) from the floating leg is $$\sum_{k=1}^n\delta_kL_{k+1}(T_k)[1+\delta_kL_{k+1}(T_k)]P(t,T_{k+1}),$$ where $$\delta_k$$ and therefore assuming $$\mathrm{d}L_{k+1}(t)=\sigma_{k+1}(t)L_{k+1}(t)\mathrm{d}W_{k+1}(t)$$, the fair value of the fixed rate should be $$R=\frac{\sum_{k=1}^n\left[\delta_kL_{k+1}(t)+\delta_k^2L_{k+1}(t)^2\exp\left(\int_t^{T_k}\sigma_{k+1}(s)^2\mathrm{d}s\right)\right]P(t,T_{k+1})}{\sum_{k=1}^n\delta_kP(t,T_k)}.$$ I was thinking that $$L_{k+1}(T_k)$$ is not a $$\mathbb{Q}^{T_k}$$-martingale, but a $$\mathbb{Q}^{T_{k+1}}$$-martingale, and so changing our numéraire pair from $$(P(\cdot,T_k),\mathbb{Q}^{T_k})$$ to $$(P(\cdot,T_{k+1}),\mathbb{Q}^{T_{k+1}})$$ should help, but I'm stuck with a floating time-$$T_k$$ payoff of $$P(t,T_{k+1})\frac{P(t,T_{k+1})}{P(t,T_k)}\mathbb{E}^{\mathbb{Q}^{T_{k+1}}}\left[\frac{\delta_kL(T_k,T_{k+1})}{P(T_k,T_{k+1})}\Bigg|\mathcal{F}_t\right],$$ which doesn't seem helpful at the slightest.

I'll only address the calculation of $$E_t^T\left[ \delta L(S,T)P(S,T)^{-1} \right].$$
$$\delta L(S,T)P(S,T)^{-1} = P(S,T)^{-2} - P(S,T)^{-1}$$
$$E_t^T\left[ P(S,T)^{-1}\right] = E_t^T\left[ 1+\delta L(S,T)\right] = 1+ \delta L(t,S,T)$$
$$E_t^T\left[ P(S,T)^{-2}\right] = E_t^T\left[ (1+ \delta L(S,T))^{2}\right]$$ $$= 1+ 2\delta L(t,S,T) + \delta^2 L(t,S,T)^2 \exp\left(\int_t^S \sigma(u)^2 du \right)$$
when $$dL(u,S,T) = \sigma(u)L(u,S,T)dW_u$$
under $$Q^T$$ measure.