Why the Standard Normal PDF Integrates to 1

The standard normal density is

f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}

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

\int_{-\infty}^{\infty} f(x)\,dx = 1
Area = 1 0
1

Start with the hard integral

Ignore the normalization constant for a moment and define

I=\int_{-\infty}^{\infty} e^{-x^2/2}\,dx

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

2

Square the integral

Introduce another independent variable, y.

I^2= \left(\int_{-\infty}^{\infty}e^{-x^2/2}\,dx\right) \left(\int_{-\infty}^{\infty}e^{-y^2/2}\,dy\right)
I^2= \iint_{\mathbb R^2}e^{-(x^2+y^2)/2}\,dx\,dy
The key observation is that x² + y² is the squared distance from the origin.
3

Switch to polar coordinates

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

r θ (x,y)

x = r cos θ

y = r sin θ

x² + y² = r²

dx dy = r dr dθ

I^2= \int_0^{2\pi} \int_0^\infty e^{-r^2/2}r\,dr\,d\theta
4

Evaluate the radial integral

Now the integral becomes simple.

\int_0^\infty e^{-r^2/2}r\,dr
Let u = r²/2, so du = r dr.
\int_0^\infty e^{-u}\,du = 1
I^2 = \int_0^{2\pi}1\,d\theta = 2\pi
I=\sqrt{2\pi}
5

Put the normalization constant back

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

\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\,dx = \frac{1}{\sqrt{2\pi}}\sqrt{2\pi}
= 1 ✓

Interview memory trick

Remember just five moves:

1
Define I
2
Square I
3
Use polar coordinates
4
Get 2π
5
Normalize