What is the rule for partial integration when we are integrating with respect to a CDF $F$ of a random variable?
For example, see this excerpt from a book:
What is going on in the second equality? In particular I don't understand how we go from $dF(x)$ to $dx$. That doesn't seem to be integration by parts, that's something that usually happens when we do substitution, no?
$$ \int\limits_u^\infty (\ln x - \ln u) \, dF(x) = -\int\limits_u^\infty (\ln x - \ln u) \, d(1-F(x)) $$ (integration by parts) $$ = -\biggl[(\ln x - \ln u) \, (1-F(x))\biggr]_u^\infty +\int\limits_u^\infty (1-F(x)) \, d(\ln x - \ln u) = \int\limits_u^\infty \frac{\overline F(x)}{x} \, dx, $$ if $\lim\limits_{x\to\infty}\ln x\cdot\overline F(x)=0$. This is fulfilled for the case under consideration: $\overline F(x)=\frac{L(x)}{x^{\alpha}}$ where $L(x)$ is some slowly varying function.
Other variant is to change order of integration: $$ \int\limits_u^\infty (\ln x - \ln u) \, dF(x) = \int\limits_u^\infty \,\left(\,\int\limits_u^x \frac1t\,dt \right)\, dF(x) $$ $\{x\in(u,\infty), u<t<x\}=\{t\in(u,\infty), x\in(t,\infty)\}$ $$ =\int\limits_u^\infty \frac1t\, \underbrace{\left(\,\int\limits_t^\infty \, dF(x)\right)}_{\overline F(t)} \,dt. $$ It is always possible to swap integrals of nonnegative functions.