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B How can I find this characteristic function

  1. Sep 8, 2016 #1
    Hello all,

    I'm trying to find the characteristic function of the random variable ##X## whose PDF is ##f_X(x)=1/(x+1)^2## where ##X\in[0,\,\infty)##. I started like this:

    [tex]\phi_X(j\nu)=E\left[e^{j\nu X}\right]=\int_0^{\infty}\frac{e^{j\nu X}}{(x+1)^2}\,dx[/tex]

    where ##j=\sqrt{-1}##. I searched the Table of Integrals book, and found the formula attached. However, according to the conditions, the real part of ##\mu## must be positive. In my case it's 0. So, I think I cannot use this formula. How then can I evaluate the above integral? I tried integration by parts by letting ##u=e^{j\nu x}## and ##dv=dx/(x+1)^2##, which gave me:

    [tex]\int_0^{\infty}\frac{e^{j\nu X}}{(x+1)^2}\,dx=\left. \frac{-e^{j\nu x}}{x+1}\right|_0^{\infty}+j\nu\int_0^{\infty}\frac{e^{j\nu x}}{x+1}\,dx[/tex]

    Is this the right way? If yes, then what? I have the same problem again in the right hand side (I mean the integral), besides the first term that I'm not sure if it's 0 or infinity at ##x=\infty##.
     

    Attached Files:

  2. jcsd
  3. Sep 12, 2016 #2

    Ssnow

    User Avatar
    Gold Member

    I think you can use a similar formular in your link, if you set in your integral ##x+1=-w## you have:

    ##=-\int_{-1}^{-\infty}\frac{e^{j\nu-j\nu w}}{w^2}dw=e^{j\nu}\int_{-\infty}^{-1}\frac{e^{-j\nu w}}{w^2}dw##

    now if you put this in wolfram Mathematica you will obtain:

    ##-\frac{j\nu w E_{i}(-j\nu w)+e^{-j\nu w}}{w}##

    as primitive. Rewrite with ##x## and take the limit ... (##E_{i}## is another expression for the exponential integral)
     
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