Lemma 1

Tag 09SB, permanent. Cite it as [Bourbaki, Tag 09SB].

Integration, IX, p. 77, § 6. Promeasures and measures on a locally convex space

For every integer $n \geqslant 0$ ,

\int_{\mathbf{R}} |x|^n e^{-x^2/2} dx = 2^{n+1 \over 2} \Gamma \left( {n+1 \over 2} \right)

(6)

\int_{\mathbf{R}} x^{2n} e^{-x^2/2} dx = (2\pi)^{1/2} \frac{(2n)!}{2^n n!}

(7)

\int_{\mathbf{R}} x^{2n+1} e^{-x^2/2} dx = 0.

(8)

Recall the formula

\Gamma(s) = \int_0^\infty u^{s-1} e^{-u} du

(9)

valid for every real number $s > 0$ (FRV, VII, §1, No. 3, Prop. 3). On making the change of variable $x = (2u)^{1/2}$ , it follows from (9) that

\int_0^\infty x^n e^{-x^2/2} dx = \int_0^\infty (2u)^{n/2} e^{-u} {1 \over 2} 2^{1/2} u^{-1/2} du = 2^{n-1 \over 2} \Gamma \left( {n+1 \over 2} \right),

whence the formula (6) since

\int_{\mathbf{R}} |x|^n e^{-x^2/2} dx = 2 \int_0^\infty x^n e^{-x^2/2} dx.

The formula (7) follows from (6) and the relation

\Gamma \left( n + \frac{1}{2} \right) = \pi^{1/2} \frac{(2n)!}{2^{2n} n!}.

(10)

For $n = 0$ , this relation reduces to $\Gamma(\frac{1}{2}) = \pi^{1/2}$ , that is, to the formula (21) of FRV, VII, §1, No. 3. The general case then follows by induction on $n$ , on taking into account the relation $\Gamma(x+1) = x \cdot \Gamma(x)$ (loc. cit., §1, No. 1).

Finally, the formula (8) follows from the fact that the function $x \mapsto x^{2n+1} e^{-x^2/2}$ is odd.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

FRV, VII, §1, No. 3 not yet on this site: FVR VII
FRV, VII, §1, No. 3, Prop. 3 not yet on this site: FVR VII
§1, No. 1