# The Gaussian Integral, a Geometrically Annotated Proof

*More* **3d graphics** and *less* symbol manipulations really make this classic proof shine.

#### Contents:

### What’s this all about

**The white area up there** is the Gaussian integral,

Aside from popping up everywhere, this integral is worth talking about because the Gaussian \(e^{-x^2}\) doesn’t have an explicit anti-derivative: calculating the integral the usual way doesn’t work (try Wolfram Alpha). So **how** do we know its value? start by thinking **outside the box**.

We collect **all the points that the spinning half bell touches** into a solid of revolution, whose top is described by the functions

Legend:

- \(r\ \ \ \ \ =\) How far along the right pointing arrow to go before taking it for a spin.
- \(e^{-r^2} =\) The height of the half bell above that point.
- \(\theta \ \ \ \ \ =\) How far to rotate the bell.

### Slice and dice

Now, the crux of the calculation is looking at **the volume** of this solid from **two different points of view**.

#### First approach

This one is common practice for solids of revolution, nothing unique here.

**Step 1:** Chop up the bell with the provided circular blades.

**Step 2:** Calculate the **area** of each slice. Look carefully and convince yourself that **each slice is a cylinder**. Also of note, if the radius of the cylinder is some value \(r\) then its height has to be \(z = e^{-r^2}\) (that’s how high the bell goes at the distance \(r\) from the origin). Conclude that

**Step 3:** Sum up the **area of all the cylinders** and you get the **volume of the 3d bell** (this is shell integration). There’s a cylinder for every radius between \(0\) and \(\infty\) (I’m showing you just a select few of them up above), so the “sum” of the areas is

While \(e^{-r^2}\) doesn’t have a nice anti-derivative, \(re^{-r^2}\) *does*, so we can calculate this integral the usual way.

Side note:these days, this part of the proof is usually obfuscated by asubstitution to polar coordinates. It’s interesting to note that the earliest renditions of this proof still had the solid of revolution front and center. See more here (item #4, partly in French).

#### Second approach

We’ll slice up the bell again, but this time we use a **different set of blades**.

It’s no coincidence that **these slices look like 2d bells themselves**, since each slice is obtained by:

- Fixing some \(x\) value (that’s the line on the floor where the blade is positioned).
- Sweeping over all possible \(y\)’s: from \(-\infty\) to \(\infty\) (that’s you moving the blade).
- Collecting all points \((x,y,z)\) between the floor and the top of the bell as we sweep.

Now, looking back at the definitions of \(x,y\) and \(z\), we get:

\[z = e^{-r^2} = e^{-(x^2+y^2)} = e^{-x^2}e^{-y^2}.\]\(r^2 = x^2+y^2\) comes from applying a trig identity.

This, combined with having \(x\) fixed at some value means that the top of each slice is indeed **the original Gaussian**, \(e^{-y^2}\), scaled down by the constant \(e^{-x^2}\).

This part only works with Gaussians, you usually won’t get such nice expressions by slicing a random solid of revolution this way. See the math here.

From here we just **follow the recipe of the first approach.**

**Step 2**: calculate the **area of each slice**. This time it’s just the scaled down area of the original 2d bell,

**Step 3:** sum all the areas. we have slices for **all possible values of** \(x\), which goes from \(-\infty\) to \(\infty\), so the sum of all of them is

Renaming \(y\) as \(x\), we finally get a second expression for the volume:

\[\left(\int_{-\infty}^{\infty}e^{-x^2}dx\right)^2.\]### Tying ends

We now know that **volume** of the 3d bell is the **square of the area** of the 2d bell. We also know from the first approach that **the volume is \(\pi\)**, so we can finish the calculation: