I'm tying myself in knots on something seemingly simple, and I'm not sure if I'm just missing something obvious. However...
Why is that:
$\int f(x) dx = F_x(x) - 1$
where does the constant $-1$ come from exactly?
By definition:
$\int_{-\infty}^{x} f(x) dx = F_x(x)$
But it is lost on me how the indefinite integral is how it is?
Thanks to anyone that can shed some light.
The proof you are reading (I assume) starts by trying to evaluate $$ \int_{-\infty}^\infty x\,f(x)\,dx $$ using integration by parts. In order to do this, we need to choose two functions $u(x)$ and $v(x)$ such that $$ x\, f(x)\,dx=u\,dv=u(x)\,v'(x)\,dx $$ The split which is most convenient for this problem is $$ u(x)=x,\qquad v'(x)=f(x) $$ Since $v'(x)=f(x)$, integrating both sides implies $$v(x)=\int f(x)\,dx +C$$ Furthermore, we know that $F_X(x)=\int_{-\infty}^x f(x)$, so $F_X(x)$ is an antiderivative of $f$. Plugging this into the above, we get $$ v(x)=F_X(x)+C $$ However, you can choose $C$ to be whatever you want. The point is, the author of the proof chose $C=-1$, because it happened to be the most convenient choice for completing the proof.
Note that the equation $$ \int f(x)\,dx =F_X(x)-1 $$ you wrote is not true! The indefinite integration symbol $\int f(x)\,dx$ refers to the entire suite of functions whose derivative is equal to $f$. What you can say is that $F_X(x)-1$ is an antiderivative of $f(x)$.