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Exponential integral

  1. Aug 14, 2010 #1

    k_c

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    1. The problem statement, all variables and given/known data

    I'm working out different pathlengths for different solids but I'm stuck on the following integral:

    2. Relevant equations

    [tex]\int[/tex] x*exp(-C*sqrt[1-x2/A2]) dx

    exp = exponential func
    sqrt = square root
    C and A are constants

    3. The attempt at a solution

    I tried to work it out with the substitution method, where u = -C*sqrt[1-x2/A2] and du = C/A2x(1-x2/A2)-1/2.

    But it seems to be getting really complex afterwards, so I was wondering if I'm overlooking the simple approach for this integral?
     
  2. jcsd
  3. Aug 15, 2010 #2
    try integration by parts.
     
  4. Aug 15, 2010 #3

    k_c

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    I tried it this way now:

    u^2 = (1-x^2/a^2)
    so du = -2x/a^2 dx and therefore x dx = -a^2/2 du

    So the integral becomes: [tex]\int[/tex] -a^2/2 exp^(-cu) du
    And then i find the solution: a^2/(2c) exp^(-cu)

    Which is a^2/(2c) exp^[-c sqrt(1-x^2/a^2)]


    I think this should be correct but the wolfram mathematica integrator gives me a different solution, so if somebody could confirm my method/solution, that would be great..
     
  5. Aug 15, 2010 #4

    ehild

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    Homework Helper
    Gold Member

    Check du.

    ehild
     
  6. Aug 15, 2010 #5

    hunt_mat

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    Homework Helper

    For the integral:
    [tex]
    \int x\exp (-c\sqrt{1-x^{2}/a^{2}})dx
    [/tex]
    I use the substitute:
    [tex]
    u^{2}=1-\frac{x^{2}}{a^{2}}
    [/tex]
    Then:
    [tex]
    xdx=a^{2}udu
    [/tex]
    This makes the integral become:
    [tex]
    a^{2}\int ue^{-u}du
    [/tex]
    This I think is easy to compute, so I will leave that part to you.
     
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