User talk:Vincent Semeria

Integral in the Gil-Pelaez theorem

Hi. I'm puzzled by this edit of yours. First, you wrote that the integral may be improper. Isn't it always improper, regardless of the univariate random variable X? Second, improper integrals are ubiquitous; is there any reason for pointing out this totally unremarkable fact?  --Lambiam 07:52, 19 January 2012 (UTC)

Hello Lambiam. This integral will be defined in Lebesgue sense when the characteristic function tends to 0 at infinity, which happens quite often, as you can see here. Besides improper integrals are rare. They are a pain to calculate too, because the usual transformation theorems don't work with them. Lastly, an improper integral is always ambiguous : you have to describe which limit you take to define it.  --Vincent 19 January 2012

Integrals of the form $\textstyle \int _{0}^{\infty }f(x)dx$ are always improper, and I don't agree such integrals are rare. It's a bit curious that you mention Lebesgue integration, since, e.g., the Dirichlet integral does not exist as a proper Lebesgue integral (as mentioned at Improper integral#Types of integrals); limits of Riemann integrals do the job here, using the standard Cauchy type of limit, implicitly understood if you don't specify otherwise.  --Lambiam 15:25, 19 January 2012 (UTC)

If you want to be more specific, you can replace "improper" by "not Lebesgue-integrable" in my comment. $\textstyle \int _{0}^{\infty }f(x)dx$ is proper, in the Lebesgue sense, when f is L1. For Gil-Pelaez, it will be L1 when the characteristic function tends to 0 at infinity. Lebesgue integration is more general than Riemann, because it allows infinite domains such as $\mathbb {R} _{+}$ (it doesn't need to take an implicit limit of finite domains). So the only case I was commenting on here, is when the charateristic function doesn't vanish at infinity. In this case yes, you will need an implicit limit of finite integrals, even with the Lebesgue integral.  --Vincent

So done.  --Lambiam 18:25, 19 January 2012 (UTC)