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AdrianMay
#3
Feb28-12, 06:49 AM
P: 98
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.