More specifically, I've come across an exercise in a book called 'Heavy Tail Phenomena' in which this kind of integral is used.

The question is, for a distribution $F$, in which $1 - F(x) \sim x^{-\alpha}, \ \ x \rightarrow \infty$ show (potentially by integrating by parts) for $\eta \geq \alpha$ that:

$$\lim_{x\rightarrow \infty} \dfrac{\int_0^x u^{\eta} F(du)}{x^{\eta}(1 - F(x))} = \dfrac{\alpha}{\alpha + \eta}$$

I do not understand what the $F(du)$ part of the integral means and I have never come across this before. Could somebody explain this to me?


1 Answer 1


Yes, from the given context, the symbol $d(F(u))$, when written along with the sign $\int$, is the symbol $F(du)$. Writing either one does not alter any truth.

For your reference, you may want to learn some Stieltjes integration. A Stieltjes integral is a number of the form $\int_{a}^{b}fdg$, which is defined as the limit of the sum $\sum_{1}^{n}f(x^{*}_{i})[g(x_{i+1}) - g(x_{i})]$ as the partitions of $[a,b]$ go finer and finer, provided only that this limit exists.

Now it would be clear that, as long as one understands a Stieltjes integral correctly, it does not matter he writing $\int_{a}^{b}f(u)d(F(u))$ or $\int_{a}^{b}f(u)F(du)$, although the latter could be not as clear as the former at the first glance. However, context usually helps!

As a related digression, in fact a Lebesgue integral also admits two common notations: $\int_{X}f(x)d\mu(x)$ or $\int_{X}f(x)\mu(dx)$. Although writing out $(x)$ seems redundant in a simple case, it would help in clarity in cases such as integrating a function of two arguments.

  • $\begingroup$ I see! I get confused between all the different types of integrals. So is it the Lebesgue-Stieltjes integral you are referring to that I ought to learn about? $\endgroup$
    – Necroticka
    Commented Jun 21, 2017 at 17:32
  • $\begingroup$ @JohnnyBreen, In a big way yes! Stieltjes integration is defined independently of Lebesgue integration. It can be shown that they are somehow related. Sort of like two persons independently choosing two starting points and ending up with arriving at the same place :) My point is that it would be better for a beginning mathematician to treat them not as one first. $\endgroup$
    – Yes
    Commented Jun 21, 2017 at 17:35

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