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Gaussian integral trouble

  1. Feb 28, 2012 #1
    Hi folks,

    I'm trying to get from the established relation:

    $$ \int_{-\infty}^{\infty} dx.x^2.e^{-\frac{1}{2}ax^2} = a^{-2}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    to the similarly derived:

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = 3a^{-4} \int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    but instead I'm winding up with:

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = (4a^{-3} - a^{-4}) \int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$.

    Evidently the difference is that I have ##a^{-3}## where I should have ##a^{-4}## but I can't seem to fault my own logic. First I differentiate the thing I started with:

    $$ -2\frac{\partial}{\partial a} [ \int_{-\infty}^{\infty} dx.x^2.e^{-\frac{1}{2}ax^2} ] = -2\frac{\partial}{\partial a} [ a^{-2}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} ] $$

    apply the chain rule:

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = -2 \{ \frac{\partial a^{-2}}{\partial a}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} + a^{-2}.\frac{\partial}{\partial a}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} \} $$

    and hit the problem in what looks like the easy bit:

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = 4a^{-3}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} - a^{-4}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    (where the last term follows from the relation I started with.)

    So where's the bug?

    Thanks in advance,
    Adrian.
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2
    I can get the correct result as well. Start with:

    $$ \int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} = \sqrt{2\pi}a^{-\frac{1}{2}} $$

    double diff right away:

    $$ 4\frac{\partial^2}{\partial a^2} \int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} = 4\frac{\partial^2}{\partial a^2} \sqrt{2\pi}a^{-\frac{1}{2}} $$

    $$ \int_{-\infty}^{\infty} dx. 4\frac{\partial^2}{\partial a^2} e^{-\frac{1}{2}ax^2} = 4\sqrt{2\pi}.\frac{-1}{2}.\frac{-3}{2}.a^{-\frac{5}{2}} $$

    $$ \int_{-\infty}^{\infty} dx. x^4. e^{-\frac{1}{2}ax^2} = 3a^{-2}\sqrt{2\pi}.a^{-\frac{1}{2}} = 3a^{-2}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    But now I'm looking at a contradiction.What was wrong with the first approach?
     
    Last edited: Feb 28, 2012
  4. Feb 28, 2012 #3
    OK, I figured it out. The starting point was supposed to be:

    $$ \int_{-\infty}^{\infty} dx.x^2.e^{-\frac{1}{2}ax^2} = a^{-1}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    so I'd get to:

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = -2 \{ \frac{\partial a^{-1}}{\partial a}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} + a^{-1}.\frac{\partial}{\partial a}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} \} $$

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = -2 \{ -a^{-2}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} + a^{-1}.\int_{-\infty}^{\infty} dx.\frac{\partial}{\partial a}e^{-\frac{1}{2}ax^2} \} $$

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = -2 \{ -a^{-2}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} + a^{-1}.-\frac{1}{2}.\int_{-\infty}^{\infty} dx.x^2.e^{-\frac{1}{2}ax^2} \} $$

    $$ \int_{-\infty}^{\infty} dx.x^4.e^{-\frac{1}{2}ax^2} = -2 \{ -a^{-2}.\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} + a^{-1}.-\frac{1}{2}.a^{-1}\int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} \} $$

    $$ = 3a^{-2} \int_{-\infty}^{\infty} dx.e^{-\frac{1}{2}ax^2} $$

    as expected. Now for that funky matrix stuff, which will doubtless lead me back here.
     
  5. Feb 28, 2012 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    One way to avoid some of the trouble would be to substitute
    y2 = ax2

    Then the power of a could be taken outside the integral.
     
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