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Most probable and average value of a function

  1. Jan 26, 2006 #1
    I've been asked to find the most probable and average energy of a neutron produced in a fission reaction. The neutron energy spectrum is represented by this function:

    \chi (E) = 0.453 e^{-1.036E} \sinh(\sqrt{2.29E})

    The first part, finding the most probable energy, seems fairly straightforward: take the derivative of the function and find the zeros - one of these zeros will be the peak of the curve, and thus the most probable energy:

    [tex]\frac{d}{dE} \chi (E)[/tex]

    I ended up with:

    [tex]\frac{d}{dE} \chi (E) = 0.453[-1.036 e^{-1.036E} \sinh(\sqrt{2.29E}) + \frac{E^{-0.5}}{2} \sqrt{2.29} cosh(\sqrt{2.29E}) e^{-1.036E}][/tex]

    which, by setting it equal to zero and manipulating, I end up with:

    [tex]\tanh(1.513 \sqrt{E}) = \frac{0.7303}{\sqrt{E}}[/tex]

    I can solve for E numerically by trying different values of E, and indeed I come up with the correct answer (~0.7) I'm just curious if it's possible to solve for E directly in this case.

    The 2nd part of the problem, finding the average energy, is giving me more trouble. The average energy should be:

    [tex]\overline{E} = \int \limits_{0} ^ {\infty} E \chi (E) \ dE[/tex]


    [tex]\overline{E} = 0.453 \int \limits_{0} ^ {\infty} E e^{-1.036E} \sinh(\sqrt{2.29E}) \ dE[/tex]

    My problem is that I can't figure out how to integrate this equation. When I try to integrate by parts, any value I choose for u and dv yields another integral that needs to be integrated by parts, which yields another etc. Can anyone give me advice on how to solve this?
    Last edited: Jan 26, 2006
  2. jcsd
  3. Jan 26, 2006 #2


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    Homework Helper

    You want to integrate something like:

    [tex]\int_0^{\infty} x e^{-x} \sinh(2a \sqrt{x} ) dx[/tex]

    Right? (the two is for later) So there will be two similar terms when you expand out the sinh. I'll only do one of them, but the other is basically the same (and I'll ignore constants that go out front):

    [tex] \int_0^{\infty} x e^{-x} e^{2a \sqrt{x}} dx = \int_0^{\infty} x e^{-x+2a \sqrt{x}} dx[/tex]

    Now use the substitution u2=x, 2udu = dx:

    [tex]\begin{align*} \int_0^{\infty} u^3 e^{-u^2+2 a u} du &= e^{a^2} \int_0^{\infty} u^3 e^{-u^2+2 a u-a^2} du \\
    &=e^{a^2} \int_0^{\infty} u^3 e^{-(u-a)^2} du \end{align*}

    At this point you can make the substitution t=u-a and expand out the (t+a)3. The odd powers of t can be integrated exactly, but for the others you'll need to look up the error function if you don't know it already. There are no exact values for their integrals on this region (-a to infinity), but there are plenty of tables you can find aproximations in. And no, your first equation has no closed form solution.
    Last edited: Jan 26, 2006
  4. Jan 26, 2006 #3

    Thanks, that's exactly what I needed. The odd values of t indeed appear easy enough to solve by making another substitution: [itex]s=t^2[/itex] I am having a bit of trouble with the even one, and I suspect it's my unfamiliarity with the error function. For example, with the integral of the form:

    [tex]\int t^2 e^{-t^2}\ ds[/tex]

    Given that the error function is defined as:

    [tex]erf(z) = \frac{2}{\sqrt{\pi}} \int \limits_{0} ^ {z} e^{-t^2} \dt[/tex]


    [tex]\int \limits_{0} ^ {z} e^{-t^2} \dt = \frac{\sqrt{\pi}}{2} erf(z)[/tex]

    so integrating by parts, letting u=[itex]t^2[/itex] and dv = [itex]e^{-t^2} dt[/itex], I end up with:

    [tex]\frac{\sqrt{\pi}*t^2}{2} erf(z) -2 \int t e^{-t^2} \dt [/tex]

    I can solve the right half using the substitution [itex]s=t^2[/itex] which reduces to:

    [tex]\frac{\sqrt{\pi}*t^2}{2} erf(z) + e^{-t^2}[/tex]

    Does that look right?
    Last edited: Jan 26, 2006
  5. Jan 26, 2006 #4


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    Homework Helper

    The way you did it, there should be an erf inside the second integral. I think a better way to do it is to use [itex]u=t[/itex] [itex]dv=t e^{-t^2} dt[/itex], so that:

    [tex]\int_0^{z} t^2 e^{-t^2} dt = -\frac{1}{2}z e^{-z^2}+\frac{1}{2}\int_0^{z} e^{-t^2} dt [/tex]
    Last edited: Jan 26, 2006
  6. Jan 26, 2006 #5
    OK, that does it. Thanks for the help.
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