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Integral of (1/(at+b))e^(-t^2)

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data

    I need to calculate the integral of (1/(at+b))e^(-t^2), that is, a negative quadratic exponential divided by a linear function at+b. I need to integrate between some positive x and +infinity.


    2. Relevant equations



    3. The attempt at a solution

    I could find the integral for b=0. In fact the integral of 1/(at) e^(-t^2) between x>0 and +infinity is equal to -1/(2a) Ei(-x^2) where Ei(x) is the exponential integral function defined as
    -Ei(-x)=E1(x)=Integral of (1/t)e^-t between x and +infinity. Since Ei can be expressed through the Incomplete Gamma function and I have the latter in Excel, in this case my problem would be - at least computationally - solved.

    However, when b is not zero I cannot find any solution. Any help guys?
     
  2. jcsd
  3. Mar 27, 2013 #2

    DrClaude

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    Staff: Mentor

    Seems like a tough one. Do you have numerical values for a, b, and x? If yes, I would suggest a numerical integration.
     
  4. Mar 28, 2013 #3
    The reason why I need an analytical solution is that I need to be able to compute the integral in a single Excel cell for any given set of parameter values. So, say that a=20 b=50 and x=80, I need to compute in a single cell (that means without being able to compute partial results in any other cell or set of cells) the integral of e^-t^2 / (20a+50) between 80 and + infinity.
    If I understand correctly what you mean by numerical integration, it seems this would require to launch a calculation with a specific software for each instance, and this is not what I need :(
    Any other ideas? Please help!
     
  5. Mar 28, 2013 #4

    DrClaude

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    Staff: Mentor

    May I ask where the integral comes from?
     
  6. Mar 28, 2013 #5
    It is the integral of the pdf of a normal distribution times 1/t. With substitution I get the integral in the title where b is the mean of the distribution. So, if anyone has a solution to compute the integral of (1/t)*PDF(NormalDistribution(mu,sigma),t) from x>0 to +infinity, this would make me very happy as well.

    Now, I could also explain where this comes from, but it is not a short story...
     
  7. Apr 2, 2013 #6
    So, in the end no one has a clue about it?
     
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