# Integration over the reals?

I'm reading this journal article and they keep on using the notation

$$\int_{R^d}$$

What does this mean, just say d=1, does it then mean

$$\int_0^{\infty}$$

or

$$\int_{-\infty}^{\infty}$$

any help much appreciated.

chiro
It means the second one, but you have to take into account the support of whatever you're integrating, but if the support is over all of R then the integral will be over all of R as well.

I am trying to find a function R(dx) in a paper by Rosinski "Tempering Stable Processes" which has the following theorem

Theorem 2.3. The Levy measure M of a tempered alpha stable distribution can be written in the form

$$M(A) = \int_{R^d}\int_0^{\infty} \textbf{I}_A(tx)t^{-\alpha-1}e^{-t}dtR(dx)$$

where I_A(tx) I assume is the indicator funtion, i.e. tx is defined on the interval A

now I'm using definition of the Gamma function kernal to say

$$\int_0^{\infty} t^{-\alpha-1}e^{-t}dt = \Gamma(-\alpha)$$

I know the Levy measure M, and so putting that in for M, I then had

$$2^{\alpha}\delta\frac{\alpha}{\Gamma(1-\alpha)}x^{-1-\alpha}e^{-0.5\gamma^{1/\alpha}x} = \Gamma(-\alpha)\int_{R^d}R(dx)$$

So I need to work out what the function R(dx) is, which is my problem. I was thinking I could just differentiate both sides which would get the integral out of the right hand side and then I could easily rearrange to find R(dx). However I am not sure about this method as I have this integral over the range R^d.

How do I work out what R(dx) is?