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$\int_a^b u(x)dF(x)$ (1)$ = u(t)F(t)|_a^b - \int_a^b F(t)u^\prime(t)dt$ (2)$ = u(b)-\int_a^b F(t)u^\prime(t)dt$

$= u(b)-(\Phi(t)u^\prime(t)|_a^b-\int_a^b \Phi(t)u^{\prime\prime}(t)dt=u(b)-\Phi(b)u^\prime(b)+\int_a^b \Phi(t)u^{\prime\prime}(t)dt$.

From (1) to (2), I know that it is a partial derivative, but shoudn't it be $u(t)F(t)|_a^b - \int_a^b F(t)u^\prime(t)dF(t)$?? How can we transform $dF(t)$ to $dt$?

I also want to know why $\Phi(b)$ is not equal to 1.

I attach the original paper in case the information is not enough. enter image description here

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That's just a question of notation. Note that

$$\mathcal{d}F(t) = f(t)\mathcal{d}t$$

Then the proof should be easy to understand.

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Your first question was addressed in another answer. For the second question (why $\Phi(b) \ne 1$), you are confusing the cdf and the density. It is true that $\int_{a}^{b}{f(t)dt}=1$ but not that $\int_{a}^{b}{F(t)dt}=1$. Think for instance of $[a,b]=[0,1]$ and the uniform distribution ($f(t)=1,F(t)=t$ for all $t \in [0,1]$).

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