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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.2012-03-06
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    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...2012-03-06
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    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.2012-03-06
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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.2012-03-06

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