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I can easily show that (substituting $\frac{x^2}{2} = t $ using the identity for Gamma function of $n+\frac{1}{2}$, then further expanding $\Gamma(n+\frac{1}{2})=\dfrac{(2n-1)!! \sqrt{\pi}}{2^n}$ and so on that

$$ I_1 =\int_{0}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx = \frac{(2n-1)!!\sqrt{\pi}}{\sqrt{2}} $$

My question is, can (and how) can I use symmetry to show that

$$ I_2 = \int_{-\infty}^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx = 2I_1 = (2n-1)!!\sqrt{2 \pi} $$

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More generally try to compute $\int_{-\infty}^{\infty} e^{- \frac{x^2}{2} } e^t$ and consider coefficients of $t$ in the result. –  Qiaochu Yuan Mar 6 '12 at 1:25
3  
You can easily do all that fancy stuff and you don't see why $\int_{-\infty}^\infty e^{-x^2/2} x^{2n}\ dx = 2 \int_0^\infty e^{-x^2/2} x^{2n}\ dx$? I'm assuming $n$ is a nonnegative integer... –  Robert Israel Mar 6 '12 at 1:31
    
Do you mean showing $f(x) = e^{-\frac{x^2}{2}}x^{2n} = e^{-\frac{(-x)^2}{2}}(-x)^{2n} = f(-x)$? It just looks too easy. –  sigma.z.1980 Mar 6 '12 at 1:37
    
Yes, it's that easy. I'm not sure where you're getting the impression the symmetry part is in any way nontrivial. –  anon Mar 6 '12 at 4:53

1 Answer 1

The below is probably as formal as it gets (substituting $-x$ for the first equality):

$$\begin{align} \int_0^\infty e^{-x^2/2}x^{2n}\,\mathrm dx &= \int_0^{-\infty} e^{-(-x)^2/2}(-x)^{2n} \cdot (-1) \,\mathrm dx\\ &= \int_{-\infty}^0 e^{-x^2/2}x^{2n}\,\mathrm dx \end{align}$$

which, together with $\displaystyle \int_{-\infty}^\infty e^{-x^2/2}x^{2n}\,\mathrm dx = \int_{-\infty}^0 e^{-x^2/2}x^{2n}\,\mathrm dx + \int_0^{\infty} e^{-x^2/2}x^{2n}\,\mathrm dx$, implies the result immediately.

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