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I One dimensional integration that Mathematica cannot do

  1. Mar 14, 2016 #1

    CAF123

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    I want to evaluate $$\int_{a+b-c}^s\,\text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon+1} (a-x)},$$ where ##a,b,c,\epsilon## are numbers, and to be treated as constants in the integration. I put this into mathematica and an hour later it is still attempting to evaluate it so I aborted the calculation. I was just wondering if someone could put the following code into Mathematica and see if it is also takes a long time on their machine?

    Integrate[(-x + a*b/c)^e/(x + c - a - b)^(e + 1)/(a - x), {x,
    a + b - c, s}, Assumptions -> e < 0 && b < 0 && a > 0]

    I tried to progress with the intergral analytically:
    Partial fractions gave a rewriting of the form $$\frac{1}{c-b}\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \left(\frac{1}{x+c-a-b} + \frac{1}{a-x}\right)$$ $$ = \frac{1}{c-b}\left(\int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon+1}} + \int_{a+b-c}^{s} \text{d}x\, \frac{(-x+ab/c)^{\epsilon}}{(x+c-a-b)^{\epsilon}} \frac{1}{a-x}\right)$$ but this probably did not help.

    Thanks!
     
    Last edited by a moderator: Mar 14, 2016
  2. jcsd
  3. Mar 15, 2016 #2

    Ssnow

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    It is improper in ##x=a+b-c## ...
     
  4. Mar 15, 2016 #3

    CAF123

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    I evaluated it initially without the factor of (a-x) in the denominator and an answer came out fine. But then I realised I had missed this piece upon typing it into Mathematica and now it takes a long time to give me an answer. ##\epsilon < 0## solves the fact it is improper at the boundary I think. Can you try putting it into your machine and seeing if is also takes a long time?
     
  5. Mar 15, 2016 #4

    Ssnow

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    yes ##\epsilon <0## solve this, I think the problem is that when you divide by ##\frac{1}{a-x}## the computation become more complex..., I tried with Mathematica but Wolfram told me false Assumptions ...
     
  6. Mar 15, 2016 #5

    Ssnow

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    You can try Taylor expanding the term ##\frac{1}{a-x}## but obviously is not an exact solution ...
     
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