The Normal Density and Its Total Area

Why the Normal Curve Needs a Normalizing Constant

A normal model is described by its center, $\mu$, and spread, $\sigma$, with $\sigma>0$. Those parameters determine the position and width of its bell-shaped curve. But a bell-shaped formula is not automatically a probability density: its total area must be exactly one. The factor in front of the normal curve is chosen to guarantee that requirement for every permitted $\mu$ and $\sigma$.

This tutorial derives that factor from the Gaussian integral. The derivation explains why the normal density contains $\sqrt{2\pi}$, rather than treating that constant as an unexplained part of a formula.

Definition: If $X$ follows a normal model with mean $\mu$ and standard deviation $\sigma>0$, written $X\sim N(\mu,\sigma^2)$, its probability density function is $$ f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right),\qquad -\infty<x<\infty. $$ The variance is $\sigma^2$, while $\sigma$ itself is the standard deviation. A probability density is nonnegative and has total integral one over its entire range.

The total-area requirement for this model is the improper integral $$ \int_{-\infty}^{\infty} f(x)\,dx=1. $$ An improper integral over the whole real line is understood as the limit of integrals over increasingly large finite intervals. Here, the normal curve is nonnegative, and its tails decrease sufficiently rapidly for this total integral to be finite.

The Gaussian Integral

The central calculation is the Gaussian integral $$ \int_{-\infty}^{\infty} e^{-u^2}\,du=\sqrt{\pi}. $$ The variable $u$ here is just an integration variable. A useful way to evaluate this one-dimensional integral is to square it, express the result as an integral over the plane, and use polar coordinates.

Let $I=\int_{-\infty}^{\infty}e^{-u^2}\,du$. This integral is positive. It is also finite: for $u\geq 1$, $u^2\geq u$, so $e^{-u^2}\leq e^{-u}$, whose tail has a finite integral; the negative tail behaves the same way. Therefore it is valid to multiply the integral by itself. Since the integrand is nonnegative and factors into a function of each coordinate, $$ I^2=\left(\int_{-\infty}^{\infty}e^{-u^2}\,du\right) \left(\int_{-\infty}^{\infty}e^{-v^2}\,dv\right) =\int_{\mathbb{R}^2}e^{-(u^2+v^2)}\,du\,dv. $$ Here, $u$ and $v$ are the two Cartesian coordinates. In polar coordinates, $u^2+v^2=r^2$, and area is represented by $r\,dr\,d\theta$, where $r\geq0$ and $0\leq\theta<2\pi$. Thus $$ I^2=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta. $$ For the inner integral, set $w=r^2$, so $dw=2r\,dr$. It follows that $$ \int_0^\infty e^{-r^2}r\,dr =\frac12\int_0^\infty e^{-w}\,dw =\frac12. $$ The last equality follows by integrating $e^{-w}$ from $0$ to a finite upper limit and then taking that limit to infinity. Consequently, $$ I^2=\int_0^{2\pi}\frac12\,d\theta =\frac12(2\pi)=\pi. $$ Because $I$ is positive, the positive square root gives $I=\sqrt{\pi}$. This establishes the Gaussian integral.

Worked Example: Evaluating the Gaussian Integral

Use the polar-coordinate argument to find the value of $\int_{-\infty}^{\infty}e^{-u^2}\,du$.

Let $I$ denote the integral. Multiplying it by itself gives a double integral over the plane: $$ I^2=\int_{\mathbb{R}^2}e^{-(u^2+v^2)}\,du\,dv. $$

In polar coordinates, the integrand is $e^{-r^2}$, and the area element is $r\,dr\,d\theta$. Therefore, $$ I^2=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta =2\pi\left(\frac12\right)=\pi. $$

Since $I$ is an integral of a positive function, $I>0$, so $I=\sqrt{\pi}$, approximately $1.772$. The positive root, rather than the negative root, is required by the integral’s sign.

Deriving the Normalizing Constant

First consider the area under the exponential part of a normal curve, before including its leading constant: $$ \int_{-\infty}^{\infty}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx. $$ Make the substitution $u=(x-\mu)/(\sqrt{2}\sigma)$. Since $\sigma>0$, this substitution increases from $-\infty$ to $\infty$ as $x$ does, and $dx=\sqrt{2}\sigma\,du$. Therefore, $$ \begin{aligned} \int_{-\infty}^{\infty}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx &=\sqrt{2}\sigma\int_{-\infty}^{\infty}e^{-u^2}\,du\\ &=\sqrt{2}\sigma\sqrt{\pi}\\ &=\sigma\sqrt{2\pi}. \end{aligned} $$ The location $\mu$ disappears because shifting the center does not change the total area. The spread $\sigma$ remains as a factor: increasing the horizontal scale by $\sigma$ increases the area under this unscaled curve by the same factor.

Now suppose the density has the form $C\exp(-(x-\mu)^2/(2\sigma^2))$, where $C$ is a constant. Its total area is $C\sigma\sqrt{2\pi}$. Requiring that area to equal one determines $C$: $$ C\sigma\sqrt{2\pi}=1 \quad\Longrightarrow\quad C=\frac{1}{\sigma\sqrt{2\pi}}. $$ This is exactly the normalizing constant in the normal density.

Theorem: For every real $\mu$ and every $\sigma>0$, $$ \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx=1. $$ The normalizing constant is the reciprocal of the area under the exponential curve before that constant is applied.

Worked Example: Checking the Standard Normal Density

The standard normal model has $\mu=0$ and $\sigma=1$, so its density is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Verify that its total area is one.

Integrate the density over the real line and use $u=x/\sqrt{2}$, giving $dx=\sqrt{2}\,du$: $$ \begin{aligned} \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx &=\frac{\sqrt{2}}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-u^2}\,du\\ &=\frac{\sqrt{2}}{\sqrt{2\pi}}\sqrt{\pi}\\ &=1. \end{aligned} $$

The exponential part alone has total area $\sqrt{2\pi}$. Multiplication by $1/\sqrt{2\pi}$ scales that area to one, as required.

Applying the Formula to Different Parameters

For any normal model, the same calculation works; only the numerical value of the normalizing constant changes with $\sigma$. The mean determines where the curve is centered but does not affect the constant. The examples below check the formula using concrete hypothetical parameter values.

Worked Example: A Normal Model With Mean 100 and Standard Deviation 12

Suppose a hypothetical measurement $X$ is modeled by $X\sim N(100,12^2)$, with measurement units called points. Write its density and verify its total area.

Here $\mu=100$ points and $\sigma=12$ points. The density is $$ f(x)=\frac{1}{12\sqrt{2\pi}}\exp\left(-\frac{(x-100)^2}{2(12)^2}\right). $$ The normalizing constant is approximately $1/(12\sqrt{2\pi})=1/30.0795=0.03325$, in reciprocal points.

For the area, substitute $u=(x-100)/(\sqrt{2}\cdot12)$, so $dx=12\sqrt{2}\,du$: $$ \begin{aligned} \int_{-\infty}^{\infty}f(x)\,dx &=\frac{12\sqrt{2}}{12\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-u^2}\,du\\ &=\frac{\sqrt{2}}{\sqrt{2\pi}}\sqrt{\pi}=1. \end{aligned} $$ The shift by 100 points does not alter the total area, and the factor involving 12 compensates for the curve’s horizontal scale.

Worked Example: Finding a Missing Constant

A proposed density for a hypothetical measurement $X$, in units, is $f(x)=C\exp(-(x-70)^2/(2\cdot9^2))$, where $C$ is an unknown positive constant. Find $C$ so the total area is one.

The unscaled exponential curve has area $\sigma\sqrt{2\pi}$, where $\sigma=9$. Its area is therefore $9\sqrt{2\pi}$, approximately $22.5597$. The area after multiplying by $C$ is $C\cdot9\sqrt{2\pi}$. Set this equal to one and solve: $$ C\cdot9\sqrt{2\pi}=1 \quad\Longrightarrow\quad C=\frac{1}{9\sqrt{2\pi}} \approx 0.04433. $$

Thus the properly normalized density is $$ f(x)=\frac{1}{9\sqrt{2\pi}} \exp\left(-\frac{(x-70)^2}{2\cdot9^2}\right). $$ The center at 70 units does not enter the calculation of $C$; the standard deviation of 9 units does.

Common Mistakes and Why Normalization Matters

A common mistake is to write the bell-shaped exponential term and assume it already has total area one. For instance, $e^{-x^2/2}$ is positive and bell-shaped, but its total area is $\sqrt{2\pi}$, not one. Multiplying by $1/\sqrt{2\pi}$ is essential. More generally, for standard deviation $\sigma$, the unscaled exponential curve has area $\sigma\sqrt{2\pi}$, so using the standard normal constant for every $\sigma$ would not work.

Wrong reasoning: “The mean only shifts the curve, so the standard normal normalizing constant works for any spread.”
Correction: A shift leaves area unchanged, but changing the horizontal scale changes area. The constant must include $1/\sigma$.
Wrong reasoning: “Any positive bell-shaped curve is a probability density.”
Correction: A density must have total area one as well as be nonnegative. If its area is not one, it must be rescaled before it is a probability density.

Normalization matters because it makes the normal curve a valid probability model for every allowed pair of parameters. The Gaussian integral supplies the specific constant that makes this possible. In the following material, the distinction between the value of a density at a point and probability areas under the curve will be treated separately.

Check Your Understanding

Use the definition and derivation to answer these questions.

What condition on $\sigma$ is required in the normal density, and why does the formula contain $\sigma\sqrt{2\pi}$ in the denominator?
In the polar-coordinate derivation, what are the limits for $r$ and $\theta$, and why does the area element contain a factor of $r$?
What is the total area under $e^{-u^2}$, and what is the total area under $\frac{1}{\sqrt{2\pi}}e^{-u^2/2}$?
A proposed curve is $C\exp(-(x-20)^2/(2\cdot5^2))$. What equation must $C$ satisfy for the curve to have total area one?
Explain why changing $\mu$ does not change the normalizing constant, whereas changing $\sigma$ does.