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Mathematica Assumptions for integration [pertaining to loop corrections]

  1. Apr 24, 2012 #1
    Long story short I have an integral which is something like this:
    (in Mathematica code)
    Integrate[1/{(1 - b^2)*{((1 - y)*z + y)^2 + (1 - y)^2*(1 - z)^2} + 2*(b^2 + 1)*{(1 - y)*(1 - z)*((1 - y)*z + y)}}, {y, 0, 1} ]

    Written without assumptions.

    b is actually only in [0,1] and I think I can limit z to [0,1] as well. z is actually in the next integral ^^; which is from 0 to 1.

    How do I add assumptions in proper syntax? I've tried two different ways and gotten VERY different results...so I'm a little concerned about my syntax and Mathematica's reference site didn't help much.

    I have been using something like ", Assumptions-> 0<=b<=1, 0<=z<=1" I think...but I'm not positive about my consistency.

    For those of you who don't have direct access to Mathematica or anything and don't read raw code super easily here's the integral in traditional form:
    [itex] \int^1_0 \frac{dy}{{(1 - b^2)*{((1 - y)*z + y)^2 + (1 - y)^2*(1 - z)^2} + 2*(b^2 + 1)*{(1 - y)*(1 - z)*((1 - y)*z + y)}}} [/itex]

    If context is helpful, the integration is the result of a Feynman parametrization of a basic QCD process. Everything is scalar though, the vectors have all been dotted or otherwise taken care of and absorbed into the b term (Sort of). The rest of the integrations are a whole other mess.
     
  2. jcsd
  3. Apr 25, 2012 #2
    Assuming[0 ≤ b ≤ 1 && 0 ≤ z ≤ 1, Integrate[1/{(1 - b^2)*{((1 - y)*z + y)^2 + (1 - y)^2*(1 - z)^2} + 2*(b^2 + 1)*{(1 - y)*(1 - z)*((1 - y)*z + y)}}, {y, 0, 1}]]

    but it looks like it is happier if you make those < rather than ≤ in all cases which I think makes sense if I look at your denominator for a few seconds.
     
  4. Apr 25, 2012 #3
    Thanks, I'll try that and see how it goes.

    Someone else recommended I use a substitution of a known function value or something for anything that's parametized like what I have?

    For example, he said that if I had some Q running from 0 to 1, I could sub {Q->Zeta[3]/3}.
    Integrate.
    Sub {Zeta[3]->3Q.

    It came out sort of close to what I had before, but I'm not entirely sure why.
    Any idea how he got this? I didn't have much of a chance to question him about it.

    I'm still playing with different factorizations to make it easier for Mathematica and myself...I'll see where that takes me, too.
     
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