Why the Standard Normal PDF Integrates to 1

The standard normal density is

f(x) = 1√(2π) e^−x²/2

To be a valid probability density function, its total area must equal 1:

∫_−∞^∞ f(x) dx = 1

Start with the hard integral

Ignore the normalization constant for a moment and define

I = ∫_−∞^∞ e^−x²/2 dx

The problem is that e^(-x²/2) has no elementary antiderivative.

Introduce another independent variable, y.

I² = (∫_−∞^∞ e^−x²/2 dx) (∫_−∞^∞ e^−y²/2 dy)

I² = &iint;_ℝ² e^{−(x²+y²)/2} dx dy

The key observation is that x² + y² is the squared distance from the origin.

Because the expression depends on x²+y², circles are the natural coordinates.

x = r cos θ

y = r sin θ

x² + y² = r²

dx dy = r dr dθ

I² = ∫₀^2π ∫₀^∞ e^−r²/2 r dr dθ

Now the integral becomes simple.

∫₀^∞ e^−r²/2 r dr

Let u = r²/2, so du = r dr.

∫₀^∞ e^−u du = 1

I² = ∫₀^2π 1 dθ = 2π

I = √(2π)

The factor 1/√(2π) is chosen precisely to cancel the Gaussian integral.

∫_−∞^∞ 1√(2π) e^−x²/2 dx = 1√(2π) · √(2π)

= 1 ✓

Remember just five moves:

1
Define I

2
Square I

3
Use polar coordinates

4
Get 2π

5
Normalize