# Another ZCB calculation

I am not getting the $$f(t,t)dt$$ term in the last equality when we have $$df(t,T)=\alpha(t,T) dt +\sigma(t,T) dW$$ and $$f(0,t)=f^{*}(0,T)$$. Instead I have an additional $$\int_{t}^{T}f(0,u)du$$ in the expression for $$X_{t}$$. Can someone see whats up here?

• Does this answer your question:quant.stackexchange.com/questions/49140/… – Magic is in the chain Oct 31 '19 at 12:15
• @Magicisinthechain not really, since $f$ is a stochastic process. – user1 Oct 31 '19 at 12:25
• I think my answer below for your another question already has it. – Gordon Oct 31 '19 at 12:26
• @Gordon I tried coping it but dont get what I want. – user1 Oct 31 '19 at 13:21