Integration over the reals?

  • Thread starter cernlife
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  • #1
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I'm reading this journal article and they keep on using the notation

[tex]\int_{R^d}[/tex]

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

[tex]\int_0^{\infty}[/tex]

or

[tex]\int_{-\infty}^{\infty}[/tex]

any help much appreciated.
 

Answers and Replies

  • #2
chiro
Science Advisor
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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.
 
  • #3
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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

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

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

[tex]\int_0^{\infty} t^{-\alpha-1}e^{-t}dt = \Gamma(-\alpha)[/tex]


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

[tex]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)[/tex]

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?
 

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