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Math Help - Integral sought for: C/x * exp(-1/2 [x-m/s]^2)

  1. #1
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    Integral sought for: C/x * exp(-1/2 [x-m/s]^2)

    I am trying to find a closed-form expression for the following integral:
     \int_{a}^{b}\frac{C}{x}exp^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx

    with 0<a<b

    Neither Maple nor Mathematica will give a ready answer.

    If I put \mu to 0, Maple gives me:

     \int\frac{C}{x}exp^{-\frac{1}{2}(\frac{x}{\sigma})^2}dx=-\frac{1}{2}*C*Ei(1,\frac{1}{2}\frac{x^2}{\sigma^2}  )


    With Ei the exponential integral. This is Ok as far as it goes, but I need the integral for x-\mu.

    If I add \mu to the function, Maple can no longer find the integral.

    I tried taking the derivative of
     -\frac{1}{2}*C*Ei(1,\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}) for clues, and that gives me \frac{C}{x-\mu} exp^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}, and the factor \frac {C}{x-\mu} is not what I'm looking for; I need \frac {C}{x}.

    Any suggestions? Am I overlooking something obvious?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Golodh View Post
    I am trying to find a closed-form expression for the following integral:
     \int_{a}^{b}\frac{C}{x}exp^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}dx

    with 0<a<b

    Neither Maple nor Mathematica will give a ready answer.

    If I put \mu to 0, Maple gives me:

     \int\frac{C}{x}exp^{-\frac{1}{2}(\frac{x}{\sigma})^2}dx=-\frac{1}{2}*C*Ei(1,\frac{1}{2}\frac{x^2}{\sigma^2}  )


    With Ei the exponential integral. This is Ok as far as it goes, but I need the integral for x-\mu.

    If I add \mu to the function, Maple can no longer find the integral.

    I tried taking the derivative of  -\frac{1}{2}*C*Ei(1,\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}) for clues, and that gives me \frac{C}{x-\mu} exp^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}, and the factor \frac {C}{x-\mu} is not what I'm looking for; I need \frac {C}{x}.

    Any suggestions? Am I overlooking something obvious?
    Why do you need this integral?

    Why not define a special function:

    G(x, \kappa)=\int_1^x \frac{e^{\frac{(\xi-\kappa)^2}{2}}}{\xi} \; d\xi

    CB
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  3. #3
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    Why not ...

    Well, the integral arises as the expectation of 1/x of a statistic with a truncated Normal distribution. If you use an ordinary non-truncated Normal distribution, the distribution of 1/x is the Cauchy distribution, which doesn't have an expectation. But I don't think that helps, and the answer seems to revolve around special functions anyway so I didn't post it in the probability and statistics thread.

    If I allow  a<0, b>0 I have to keep track of the singularity of 1/x in 0, which I'm not sure how to do. This work has already been done for the case that \mu=0 in the investigation of the Ei function,which isn't something I know how to replicate.

    So why not define a special function like e.g. G(x,\kappa) and go with that?

    Well, for two reasons.

    First of all, the integral seems tantalizingly close to one that's already expressible in terms of a known special function (Ei). If I'm to define a "new" special function, I should have some confidence I'm not overlooking some obvious trick that will reduce it to a known special function, right. I have no idea how to go about that. Any suggestions on this part?

    If I can be reasonable certain that my integral isn't expressible in known elementary and special functions I'm happy to define a new special function G(x,\kappa) as you suggest and do a simple investigation of its properties (extremes, inflexion points, limits).

    Secondly, I can get numerical values for the integral I'm looking for from Matlab, but that just gives me numbers but not a lot of insight. Besides I can't readily check the numbers if I do, so I'd either have to trust Maple to get it right, or program the function in a dedicated numerical program like Matlab, Scilab or Octave and then use a numerical integration algorithm that I know to be correct to check on Maple's numbers. I will if I have to, but again, I want to be sure that it's not redundant (i.e. that G(x,\kappa) isn't some simple expression of documented quantities).
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Golodh View Post
    Well, the integral arises as the expectation of 1/x of a statistic with a truncated Normal distribution. If you use an ordinary non-truncated Normal distribution, the distribution of 1/x is the Cauchy distribution, which doesn't have an expectation. But I don't think that helps, and the answer seems to revolve around special functions anyway so I didn't post it in the probability and statistics thread.
    The problem about the mean of 1/x with the non-truncated normal distribution is due to the singularity at x=0 not the tails, so if your integration interval includes 0, you will have the same problem.

    CB
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  5. #5
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    Truncated ...

    The problem about the mean of with the non-truncated normal distribution is due to the singularity at not the tails, so if your integration interval includes , you will have the same problem.
    Of course, which is why I'm interested in the truncated Normal (so that I can exclude 0 from the integration domain). In the question of the integral as posed the problem of the singularity for x=0 plays no role.

    The digression about the singularity and the expectation of was by way of background information, which I left out because I felt it might obscure the real problem,which is how to deal with the integral.
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