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$$\int_{-\infty}^{\infty} x^2e^{-x^2} \,dx$$

$g(x) = x^2e^{-x^2}$

Well, After computing it's fourier transform, which is $g(w) =\frac{2-w^2}{8\sqrt\pi}\cdot e^{\frac{-w^2}{4}}$.

In the solution they used some formula and said that:

$\int_{-\infty}^{\infty} x^2e^{-x^2} \,dx = 2\pi g(0) = \frac{\sqrt\pi}{2}$.

Well I don't understand what formula they used, and why did they make it $w = 0$. I thought about the $g(x) = \int_{-\infty}^{\infty} g(w)e^{iwx} dw$ and $w=0$. which didn't quite work..

Edit: I think I solved it. used the definiton $\frac{1}{2\pi}\int_{-\infty}^{\infty} g(x)e^{iwx} dx = g(w)$

3 Answers 3

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For the undefinite integral, use integration by parts:

$$\mathcal{I}\left(x\right)=\int x^2e^{-x^2}\space\text{d}x=-\frac{xe^{-x^2}}{2}+\frac{1}{2}\int e^{-x^2}\space\text{d}x\tag1$$

Well, we have that:

$$\int\frac{2e^{-x^2}}{\sqrt{\pi}}\space\text{d}x=\frac{2}{\sqrt{\pi}}\int e^{-x^2}\space\text{d}x=\text{erf}\left(x\right)+\text{C}\tag2$$

So, we get:

$$\mathcal{I}\left(x\right)=\frac{\sqrt{\pi}\text{erf}\left(x\right)-2xe^{-x^2}}{4}+\text{C}\tag3$$

Now, for the boundaries we get:

  1. $$\lim_{x\to-\infty}\mathcal{I}\left(x\right)=-\frac{\sqrt{\pi}}{4}\tag4$$
  2. $$\lim_{x\to\infty}\mathcal{I}\left(x\right)=\frac{\sqrt{\pi}}{4}\tag5$$
  • 0
    Hey. look at my edit. – 2017-02-05
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    I used the definiton $\frac{1}{2\pi}\int_{-\infty}^{\infty} g(x)e^{iwx} dx = g(w)$** – 2017-02-05
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    @IlanAizelmanWS As you see in my answer, you don't have to use your 'definition' formula because using my method it also works! – 2017-02-05
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    Too long for 10 points in an exam question when the first was to compute its transform. – 2017-02-05
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    @IlanAizelmanWS Maybe that is true, but your method is not wrong! – 2017-02-05
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By integration by parts $$ \int_{-\infty}^{+\infty}x^2 e^{-x^2}\,dx = \int_{-\infty}^{+\infty}\frac{x}{2}\left(2x e^{-x^2}\right)\,dx = \frac{1}{2}\color{red}{\int_{-\infty}^{+\infty}e^{-x^2}\,dx} = \frac{\color{red}{\sqrt{\pi}}}{2}$$ So it is enough to know the gaussian integral, there is no need to invoke Fourier transforms or the $\Gamma$ function.

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Just for fun, here's a method that completely sidesteps integration by parts.

\begin{align} \int_{-\infty}^\infty x^2e^{-x^2}dx&=-\int_{-\infty}^\infty\frac{\partial}{\partial\mu}e^{-\mu x^2}dx\bigg\vert_{\mu=1}\\ &=-\frac{d}{d\mu}\int_{-\infty}^\infty e^{-\mu x^2}dx\bigg\vert_{\mu=1}\\ &=-\sqrt{\pi}\frac{d}{d\mu}\left(\frac{1}{\sqrt{\mu}}\right)\bigg\vert_{\mu=1}\\ &=\frac{\sqrt{\pi}}{2\mu^{3/2}}\bigg\vert_{\mu=1}\\ &=\frac{\sqrt{\pi}}{2} \end{align}

As with Jack's method, you only need to know the Gaussian integral.