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Mathematica Tough Log Integration

  1. Mar 10, 2010 #1

    Hepth

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    Gold Member

    I have an integration to perform

    [tex]\sqrt{\text{E1}^2-m^2} Log\left( \frac{m^2-m_B E1-m_B\sqrt{E1^2-m^2}}{m^2-m_B E1+m_B \sqrt{E1^2-m^2}}\right)[/tex]
    Code (Text):
    Sqrt[E1^2 - m^2]
      log((m^2 - mB (Sqrt[E1^2 - m^2] + E1))/(
      mB (Sqrt[E1^2 - m^2] - E1) + m^2))
    over the region {E1,m,mB/2}

    E1 is the variable. m is the minimum. mB/2 the max.

    The integrand is valid over that region, and the two limits are both the points where the integrand crosses the axis.

    I'd like an algebraic solution if possible. I can always go to numerical integration, but I'd prefer not to at this stage.

    Mathematica won't solve it without limits, it just spits out the integral again. Same thing with the limits, but takes about 5 minutes to spit out the input.

    Anyone have any non-numerical ideas?


    the plot:
    Code (Text):

    tmp2 = Sqrt[E1^2 - m^2]
       Log[(m^2 - mB (Sqrt[E1^2 - m^2] + E1))/(
       mB (Sqrt[E1^2 - m^2] - E1) + m^2)];
    tmp3 = tmp2 /. {mB -> 5000., m -> 100., \[Lambda]2 -> 1.} //
      FullSimplify;
    Plot[tmp3, {E1, 100, 6000}, PlotRange -> {{0, 3000}, {0, 40000}}]
     
    Last edited: Mar 11, 2010
  2. jcsd
  3. Mar 10, 2010 #2

    Hepth

    User Avatar
    Gold Member

    I think my limits are too high... nevermind.

    EDIT: no theyre not, :) please help
     
    Last edited: Mar 10, 2010
  4. Mar 10, 2010 #3
    Have you attempted to simplify the integrand by hand?
     
  5. Mar 10, 2010 #4

    Hepth

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    Gold Member

    yeah, it simplifies a little, but doesn't change the problem. I think I made a mistake somewhere though, so give me another day before trying this. It'll probably change.
     
  6. Mar 11, 2010 #5

    Hepth

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    Gold Member

    I simplified it some. Still a problem. Original equation updated.

    The integration bounds are the same as the functions limits in the positive E1 region.
     
  7. Mar 11, 2010 #6

    Dale

    Staff: Mentor

    Usually, if Mathematica cannot solve the integral then that means that there is no analytical solution. I.e. Mathematica can do pretty much any integral that you would find in a book.

    The only case where I have found a problem where an integral would not solve and I could do something about it was when there was some coordinate transformation that I could make which simplified it into something that was solvable.
     
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