Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Gaussian integral

Theorem: The definite integral of $\mathrm{exp}\left[ -x^2 \right]$ from $-\infty$ to $+\infty$ is equal to the square root of $\pi$:

$\label{eq:norm-gi} \int_{-\infty}^{+\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x = \sqrt{\pi} \; .$

Proof: Let

$\label{eq:I} I = \int_{0}^{\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x$

and

$\label{eq:IP} I_P = \int_{0}^{P} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x = \int_{0}^{P} \mathrm{exp}\left[ -y^2 \right] \, \mathrm{d}y \; .$

Then, we have

$\label{eq:IP-I} \lim\limits_{P \rightarrow \infty} I_P = I$

and

$\label{eq:IP2-I2} \lim\limits_{P \rightarrow \infty} I_P^2 = I^2 \; .$

Moreover, we can write

$\label{eq:IP2} \begin{split} I_P^2 &\overset{\eqref{eq:IP}}{=} \left( \int_{0}^{P} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x \right) \left( \int_{0}^{P} \mathrm{exp}\left[ -y^2 \right] \, \mathrm{d}y \right) \\ &= \int_{0}^{P} \int_{0}^{P} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y \\ &= \iint_{S_P} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y \end{split}$

where $S_P$ is the square with corners $(0,0)$, $(0,P)$, $(P,P)$ and $(P,0)$. For this integral, we can write down the following inequality

$\label{eq:IP2-ineq} \iint_{C_1} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y \leq I_P^2 \leq \iint_{C_2} \mathrm{exp}\left[ - \left( x^2 + y^2 \right) \right] \, \mathrm{d}x \, \mathrm{d}y$

where $C_1$ and $C_2$ are the regions in the first quadrant bounded by circles with center at $(0,0)$ and going through the points $(0,P)$ and $(P,P)$, respectively. The radii of these two circles are $r_1 = \sqrt{P^2} = P$ and $r_2 = \sqrt{2 P^2} = P \sqrt{2}$, such that we can rewrite equation \eqref{eq:IP2-ineq} using polar coordinates as

$\label{eq:IP2-ineq-PC} \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_1} \mathrm{exp}\left[ -r^2 \right] \, r \, \mathrm{d}r \, \mathrm{d}\theta \leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_2} \mathrm{exp}\left[ -r^2 \right] \, r \, \mathrm{d}r \, \mathrm{d}\theta \; .$

Solving the definite integrals yields:

$\label{eq:IP2-ineq-PC-int} \begin{split} \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_1} \mathrm{exp}\left[ -r^2 \right] r \, \mathrm{d}r \, \mathrm{d}\theta &\leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \int_{0}^{r_2} \mathrm{exp}\left[ -r^2 \right] r \, \mathrm{d}r \, \mathrm{d}\theta \\ \int_{0}^{\frac{\pi}{2}} \left[ -\frac{1}{2} \mathrm{exp}\left[ -r^2 \right] \right]_{0}^{r_1} \, \mathrm{d}\theta &\leq I_P^2 \leq \int_{0}^{\frac{\pi}{2}} \left[ -\frac{1}{2} \mathrm{exp}\left[ -r^2 \right] \right]_{0}^{r_2} \, \mathrm{d}\theta \\ -\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( \mathrm{exp}\left[ -r_1^2 \right] - 1 \right) \, \mathrm{d}\theta &\leq I_P^2 \leq -\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( \mathrm{exp}\left[ -r_2^2 \right] - 1 \right) \, \mathrm{d}\theta \\ -\frac{1}{2} \left[ \left( \mathrm{exp}\left[ -r_1^2 \right] - 1 \right) \theta \right]_{0}^{\frac{\pi}{2}} &\leq I_P^2 \leq -\frac{1}{2} \left[ \left( \mathrm{exp}\left[ -r_2^2 \right] - 1 \right) \theta \right]_{0}^{\frac{\pi}{2}} \\ \frac{1}{2} \left( 1 - \mathrm{exp}\left[ -r_1^2 \right] \right) \frac{\pi}{2} &\leq I_P^2 \leq \frac{1}{2} \left( 1 - \mathrm{exp}\left[ -r_2^2 \right] \right) \frac{\pi}{2} \\ \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -P^2 \right] \right) &\leq I_P^2 \leq \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -2 P^2 \right] \right) \end{split}$

Calculating the limit for $P \rightarrow \infty$, we obtain

$\label{eq:IP2-ineq-PC-int-lim} \begin{split} \lim\limits_{P \rightarrow \infty} \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -P^2 \right] \right) \leq \lim\limits_{P \rightarrow \infty} I_P^2 &\leq \lim\limits_{P \rightarrow \infty} \frac{\pi}{4} \left( 1 - \mathrm{exp}\left[ -2 P^2 \right] \right) \\ \frac{\pi}{4} \leq I^2 &\leq \frac{\pi}{4} \; , \end{split}$

such that we have a preliminary result for $I$:

$\label{eq:I-qed} I^2 = \frac{\pi}{4} \quad \Rightarrow \quad I = \frac{\sqrt{\pi}}{2} \; .$

Because the integrand in \eqref{eq:norm-gi} is an even function, we can calculate the final result as follows:

$\label{eq:norm-gi-qed} \begin{split} \int_{-\infty}^{+\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x &= 2 \int_{0}^{\infty} \mathrm{exp}\left[ -x^2 \right] \, \mathrm{d}x \\ &\overset{\eqref{eq:I-qed}}{=} 2 \, \frac{\sqrt{\pi}}{2} \\ &= \sqrt{\pi} \; . \end{split}$
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Metadata: ID: P196 | shortcut: norm-gi | author: JoramSoch | date: 2020-11-25, 04:47.