Why Does My Calculation of the Gaussian Integral for x^4 Differ?

[AdrianMay]

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.

 

[AdrianMay]

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?

 

[AdrianMay]

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.

 

[mathman]

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.