Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
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
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.