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How to calculate this integral: $$\int_{-\infty}^{\infty} x^2 \mathrm e^{-\frac{1}{2}x^2} \; \mathrm dx$$

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    What have you done so far? Any thoughts on how might tackle it? – 2017-02-16
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    Should the lower bounds be negative... – 2017-02-16
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    You sure there is no typo? Also have you heard of differentiating under the integral sign? – 2017-02-16
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    Soryy I made a mistake @Chinny84 – 2017-02-16

3 Answers 3

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Hint: we have the Gaussian integral:

$$f(t)=\int_{-\infty}^{+\infty}e^{-tx^2}\ dx=\sqrt{\frac\pi t}$$

We then have

$$-f'(1/2)=\int_{-\infty}^{+\infty}x^2e^{-\frac12x^2}\ dx=\dots$$

Can you finish the rest?

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Hint: Write,

$$I=\int_{-\infty}^{\infty} x^2e^{-\frac{1}{2}x^2}dx>0$$

$$I=\int_{-\infty}^{\infty} y^2e^{-\frac{1}{2}y^2} dy$$

So that,

$$I^2=\iint_{\mathbb{R}^2} x^2y^2e^{-\frac{1}{2}(x^2+y^2)} dA$$

Converting to polar we have,

$$I^2=\int_{0}^{2\pi} \int_{0}^{\infty} r^5\cos^2 (\theta) \sin^2 (\theta)e^{-\frac{1}{2}r^2} dr d \theta$$

$$I^2=\int_{0}^{2\pi} \cos^2 (\theta) \sin^2 (\theta) d\theta \int_{0}^{\infty} r^5e^{-\frac{1}{2}r^2} dr$$

$$I^2=\int_{0}^{2\pi} \frac{1}{4} \sin^2 (2\theta) d\theta \int_{0}^{\infty} r^5e^{-\frac{1}{2}r^2} dr$$

On the second integral try $u=r^2$ and integration by parts. Or $u=\frac{1}{2}r^2$ and remember the integral form of the gamma function.

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    Good 'ole fashion polar methods. :-) – 2017-02-16
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    how did you get from first set of equations to the third ? – 2017-02-16
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    @Chinny84 are you familiar with polar transformations? That's all this is. – 2017-02-16
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    @SimplyBeautifulArt yup I am. But Ahmed had $$\int (x^2+y^2)\mathrm{e}^{\alpha(x^2+y^2)}dxdy$$ rather than $$\int (x^2y^2)\mathrm{e}^{\alpha(x^2+y^2)}dxdy$$ which he has now. – 2017-02-16
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    Sorry. ${}{}{}$ – 2017-02-16
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Let $\dfrac{1}{2}x^2=u$ then $$\int_{-\infty}^{\infty} x^2 \mathrm e^{-\frac{1}{2}x^2} \; \mathrm dx=2\int_{0}^{\infty} 2u \mathrm e^{-u} \dfrac{du}{\sqrt{2u}}=2\sqrt{2}\int_{0}^{\infty} u^\frac12 \mathrm e^{-u} du=2\sqrt{2}\Gamma(\frac32)=\sqrt{2\pi}$$