I was studying a proof where the author shows that if the range of x is $\mathbb R_+$ and $F$ is the cumulative distribution function then:
$$E[X] = \int_{0}^\infty (1-F(x))dx $$
However on one part of the proof, when integrating by parts, he states that if
$$ dv = -f(x)$$ then $$ v = 1-F(x)$$
I do know that $$ F(a)= \int_{-\infty}^a f(x) dx$$
but I can´t find a way to understand why the $-1$ of in the expression of $v$.
Can anyone help me out?
Here is one way of doing it, but you need to be able to justify an interchange of an iterated integral.
You have $F(x) = \int_0^x f(x) dx$ and $EX = \int_0^\infty x f(x) dx$.
Write $x = \int_0^x dt$ to get $EX = \int_{x=0}^\infty \int_{t=0}^x f(x) dtdx$. All the quantities are positive, so we can switch the order of integration to get $EX = \int_{t=0}^\infty \int_{x=t}^\infty f(x)dx dt$. Since we have $F(t)+ \int_{x=t}^\infty f(x)dx dt = 1$, we get $EX = \int_{t=0}^\infty (1-F(t))dt$, which is the desired result.