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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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    Yes, it's that easy. I'm not sure where you're getting the impression the symmetry part is in any way nontrivial.2019-04-27

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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.