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Solution to the integral,i.e, expected value of a function of normal variable

  1. Jun 26, 2012 #1
    I want to calculate [itex]\int_a^b \frac{1}{\sqrt{2 \pi \sigma^2}} e^{(-(x-\mu)/\sigma^2)} log_2 (1 + e^{-x}) dx[/itex]
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  3. Jun 26, 2012 #2

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    You'll need to use numerical techniques. The above integral doesn't have a closed form solution in the elementary functions.
  4. Jun 26, 2012 #3
    can any approximation be made?
  5. Jun 26, 2012 #4


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    Hey ait.abd and welcome to the forums.

    Numerical techniques are ones that give approximate answers in general. You can supply parameters to get a good enough approximation (like for example a number good enough to say 4 decimal places) for the better implementations.

    If you are unsure, just use a common package for numerical calculation.

    You should probably try searching online for a numeric integrator Java applet, or go to www.wolframalpha.com and enter in your expression to get an approximate answer.
  6. Jun 26, 2012 #5
    Thanks chiro. But, wolfram online integrator doesn't work for this expression as it tries to compute the exact expression. I can perform numerical integration but I want answer in terms of $a$ and $b$. Numerical integration will calculate the answer for a particular $a$ and $b$.
  7. Jun 26, 2012 #6


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    If you need something analytical, you're going to have to develop some sort of approximation for the integral for certain parameter regimes. For example, for [itex]\sigma \rightarrow 0[/itex], the Gaussian essentially becomes a delta function and you would get

    [tex]\log_2(1+e^{-\mu})\left[\Theta(b-\mu)+\Theta(\mu - a)\right][/tex]
    as the result (the step functions [itex]\Theta[/itex] guarantee that mu is between a and b). So, for [itex]\sigma[/itex] small, a rather crude approximation might be

    [tex]\log_2(1+e^{-\mu}) \int_a^b \frac{dx}{\sqrt{2\pi}\sigma} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right],[/tex]
    where the integral can be evaluated in terms of the error function. There is a more systematic way to generate this approximation called the method of steepest descent. (You'll have to look that up in a book; I'm afraid the wikipedia article isn't very helpful).

    You might also be able to write down an infinite series for the integral. However, when I tried this by expanding the logarithm in powers of e^(-x), I got a sum which looks like it doesn't converge, indicating that either I made a mistake in my calculation or that switching the integral and sum isn't valid in this case.
  8. Jun 26, 2012 #7
    Just looking at the form of equation. It seems that complex contour integral MAY work.
  9. Jun 26, 2012 #8


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    The integral in the original post does NOT involve [itex]e^{-x^2}[/itex]. It is, rather, of the form [itex]e^{-x}[/itex].
  10. Jun 26, 2012 #9

    D H

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    I suspect that that may have been a typo. Even if it is a typo, it doesn't help. Either way ( exp(-x2) vs exp(-x) ), this function is not integrable in the elementary functions. Since it's not an integral that is widely used, it's dubious that someone has come up with a nifty way to evaluate it.

    To ait.abd: You need to learn how to do numerical integration sometime. If this is the right integral, that sometime is now.
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