Lemma 2

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

For every complex number $y$ ,

(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{ixy} dx = e^{-y^2/2}.

(11)

In particular,

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

The change of variable $x \mapsto -x$ yields

(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{i xy} dx = (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{-ixy} dx;

since $\cos u = \frac{e^{iu} + e^{-iu}}{2}$ for every complex number $u$ , it follows that

(12) \quad (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} e^{i xy} dx = (2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} \cos xy \, dx .

For every integer $n \geqslant 0$ , set

g_n(x) = (-1)^n (2\pi)^{-1/2} \frac{(xy)^{2n}}{(2n)!} e^{-x^2/2}.

By (7),

(13) \quad \int_{\mathbf{R}} |g_n(x)| \, dx = \frac{1}{n!} \left( \frac{|y|^2}{2} \right)^n

(14) \quad \int_{\mathbf{R}} g_n(x) \, dx = \frac{1}{n!} \left( -\frac{y^2}{2} \right)^n,

whence

\sum_{n=0}^{\infty} \int_{\mathbf{R}} |g_n(x)| \, dx = e^{|y|^2/2} < +\infty.

Since, moreover,

(2\pi)^{-1/2} e^{-x^2/2} \cos xy = \sum_{n=0}^{\infty} g_n(x),

this equality can be integrated term by term, whence

(2\pi)^{-1/2} \int_{\mathbf{R}} e^{-x^2/2} \cos xy \, dx = \sum_{n=0}^{\infty} \int_{\mathbf{R}} g_n(x) \, dx = e^{-y^2/2}

by (14). The formula (11) then follows from (12).

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

Cited by

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