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B How can I find this integral?

  1. Sep 10, 2016 #1
    Hi all,

    I have this integral:

    [tex]\int_0^{\infty}\nu^{n}e^{-j\nu[y+n]}\left[E_1(-j\nu)\right]^n\,d\nu[/tex]

    How can I evaluate this integral? I found the attached integral formula in the table of integrals, but the exponential integral function ##E_1(.)## is not raised to the power of an exponent ##n\ge 0##. Any idea?

    Thanks in advance
     

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  3. Sep 11, 2016 #2

    Ssnow

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    Hi, to find an exact answer seems hard! especially for every ##n##. I suggest you to try with a math software as Mathematica or use some approximation of ##E_{1}(x)## as the following:

    ## E_{1}(x)=\int_{x}^{+\infty}\frac{e^{-t}}{t}dt \sim e^{-x}\sum_{n=0}^{+\infty}(-1)^{n}\frac{n!}{x^{n+1}}##

    that is a good approximation for ##x\rightarrow +\infty##.
     
  4. Sep 12, 2016 #3
    I could use numerical integration I guess, but this isn't the final result in my analysis. The infinite series approximation is still problematic as it's raised to the power and its upper limit is infinite.
     
  5. Sep 12, 2016 #4
    Well, I suggest you use integration by parts since

    $$\frac {\partial }{\partial x} E(-xy)^n =-n\frac {e^{xy}}{x} E(-xy)^{n-1} $$

    I think it is easy for small values of n. Try to generalize it.
     
  6. Sep 12, 2016 #5
    Thanks. Do you have a resource for this formula?
     
  7. Sep 12, 2016 #6
    It is a simple differentiation of power

    $$\frac {d}{dx}[g (x)]^n = n g'(x) [g (x)]^{n-1}$$
     
  8. Sep 12, 2016 #7
    Right. But how to differentiate ##E_1(-j\nu)##. I know that

    [tex]E_1(-j\nu)=\int_1^{\infty}\frac{e^{j\nu t}}{t}\,dt[/tex].

    This means that
    [tex]\frac{\partial}{\partial \nu}E_1(-j\nu)=\int_1^{\infty}\frac{\partial}{\partial \nu} \frac{e^{j\nu t}}{t}\,dt=j\int_1^{\infty}e^{j\nu t}\,dt[/tex]

    Right? How did you get your expression?

    Thanks
     
  9. Sep 12, 2016 #8
    Am I right in this or not?
     
  10. Sep 13, 2016 #9

    Ssnow

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    hhmmm, it is convenient to have the integral in the form ##E_{1}(-j\nu)=\int_{-j\nu}^{+\infty}\frac{e^{-t}}{t}dt## so you can apply the fundament theorem of calculus when you derive respect ##\nu##...
    To be precise it is better take the limit:

    ##\lim_{b\rightarrow +\infty}\frac{\partial}{\partial\nu}\int_{-j\nu}^{b}\frac{e^{-t}}{t}dt =... ##
     
  11. Sep 16, 2016 #10
    Yes as Ssnow suggested. Can you try sovle the integral for n=2 ?

    Sorry I for the late response I was travelling.
     
    Last edited: Sep 16, 2016
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