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Integral of Gaussian function, for squared x

  1. Jul 27, 2014 #1
    1. The problem statement, all variables and given/known data
    I am trying to compute an integral, as part of the expected value formula (using a Gaussian PDF)

    [tex]\int_{-∞}^{∞} (x)^2 p(x) dx [/tex]

    Where p(x) is the Gaussian probability density function:
    [tex]\frac{1}{\sigma \sqrt(2 \pi)} \exp(\frac{-x^2}{2 \sigma^2})[/tex]

    My aim after this is to be able to compute for all even x^n in the above formula. For all odd x^n, the positive and negative componets cancel out, with an computation of zero for all odd functions.


    2. Relevant equations
    Wikipedia lists two equations that relate to this:
    [1]https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions

    [1]
    [tex]\int_{-∞}^{∞} (x)^2 \phi(x)^n dx = \frac{1}{\sqrt(n^3 (2 \pi)^{n-1})} [/tex]

    and
    [2]https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

    [2]
    The range here is zero to infinity, however as this function is even, the result for minus infinity to infinity should be twice what the below computes.
    [tex]\int_{0}^{∞} (x)^n \exp(-\alpha x^2) dx= \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})[/tex]
    Where
    [tex]n=2 k, k integer, \alpha > 0[/tex]

    3. The attempt at a solution
    Using [1], I can compute a answer equalling 1, which is apparently the right answer. However when using [2] (where k=1), I compute a different anwser:

    Compute:
    Let [tex]\alpha = \frac{1}{2 \sigma^2}[/tex]
    and
    [tex]k=1[/tex]

    then:
    [tex]\int_{0}^{∞} (x)^2 \exp(-\alpha x^2) dx = \frac{(2 k -1)!!}{2^{k+1} \alpha^k} \sqrt(\frac{\pi}{\alpha})[/tex]

    [tex]=\frac{1!!}{4 \alpha}\sqrt(\frac{\pi}{\alpha})[/tex]

    [tex]=\frac{\sigma^2}{2}\sqrt(2 \sigma^2 \pi)[/tex]

    [tex]=\frac{\sigma^2}{2} \sigma \sqrt(2 \pi)[/tex]

    [tex]=\frac{\sigma^3 \sqrt(2 \pi)}{2} [/tex]

    Now, this is where I think I messed up. I previously removed
    [tex]\frac{1}{\sigma \sqrt(2 \pi)}[/tex]
    from inside the integral, and attempted to multiply it by the obtained result (and then doubled everything, as it is an even function)

    [tex]2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2} [/tex]

    Then by cancelling:
    [tex]=\sigma [/tex]

    I'm thinking this is because I took part of the PDF out of the integral, and then changed the limits. However, I'm not sure how to go about working this out while leaving it in there.

    I will keep browsing around for a solution to this problem (and the more general x^n), and would really appreciate any hints on this.

    Thanks.
     
  2. jcsd
  3. Jul 27, 2014 #2

    Ray Vickson

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    First get
    [tex] I_n = \int_0^{\infty} x^n \phi(x) \, dx, \:\: \phi(x) =
    \frac{1}{\sqrt{2 \pi}} e^{-x^2/2},[/tex]
    then obtain the general result by re-scaling the variable. Note that ##x \phi(x) \, dx = d(-\phi(x))## so we can use integration by parts:
    [tex] \int x^n \phi(x) \, dx = \int u dv,\\
    u = x^{n-1}, \: dv = x \phi(x) \, dx = d(-\phi(x))[/tex]
     
  4. Jul 27, 2014 #3

    Orodruin

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    Alternatively, change variables to ##t = x^2## and use the definition of the Gamma function.
     
  5. Jul 28, 2014 #4
    Thank you for your suggestions.

    Ray, I will attempt to work it out using your way later today, and see how it goes.

    Orodruin, I have had a go at your method, but have not had much luck.

    [tex] t = x^2[/tex]

    [tex] \Gamma(z) = \int_0^{∞} t^{z-1} exp(-t) dt [/tex]

    [tex] \Gamma(3) = \int_0^{∞} t^{3-1} \exp(-t) dt [/tex]

    [tex] \Gamma(3) = (3-1)! = 2 [/tex]

    Then for the range [itex]{-∞,0}[/itex], this answer should be the same, so I get an answer of 4 overall.I'm working under the assumption that the answer should be 1
     
    Last edited: Jul 28, 2014
  6. Jul 28, 2014 #5

    Orodruin

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    Well, the answer cannot be 1. It should be the variance of the distribution (since the expectation value is zero). In fact, you had it (almost) right in your first post... Also note that in the substitution, z is not 3. If you do it correctly, z should have a half-integer value (which is why you end up with a semi-factorial and 2^(k+1)).

    Edit: Well, of course the answer would be one if the variance is ... :)
     
  7. Jul 28, 2014 #6
    Ok, I see what you mean about the variance, the definition of [itex]\phi[/itex] on the wiki page assumes [itex]\sigma=1[/itex]. Using [itex]\sigma=1[/itex], my calculation did evaluate to one, which is where I must have gotten that assumption from.

    I redid my calculations with including [itex]\sigma[/itex], and realised I made a silly error in my cancellation.
    [tex]2 \frac{1}{\sigma \sqrt(2 \pi)} \frac{\sigma^3 \sqrt(2 \pi)}{2} [/tex]

    Should evaluate to [itex]\sigma^2[/itex], not [itex]\sigma[/itex].

    I will look more into the Gamma function as well, and try to understand it from that direction. Thanks
     
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