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The Gaussian — √π from a Bell Curve

The bell curve e^(−x²) has no closed-form antiderivative. Yet the area underneath it from −∞ to ∞ is exactly √π. The proof, found by Poisson, uses no calculus you don't already know — just one trick of geometric symmetry.

The bell curve and its running area
∫−∞∞ e−x² dx = √π
limit ±L
3.00
area in [−L, L]
1.7724
√π
1.7725
The square trick — Poisson's proof
I² = ∫∫ e−(x²+y²) dx dy = ∫₀∞ e−r² · 2πr dr = π
2D integral
3.1413
π
3.1416
→ I = √π
1.7725
limit L 3.00
why squaring it makes π appear
Call the answer I. Then I² is a two-dimensional integral:

I² = (∫ e^(−x²) dx) · (∫ e^(−y²) dy) = ∫∫ e^(−(x²+y²)) dx dy

The integrand e^(−(x²+y²)) = e^(−r²) is rotationally symmetric. Switch to polar coordinates: dx dy → r dr dθ:

I² = ∫₀^(2π) ∫₀^∞ e^(−r²) · r dr dθ = 2π · ∫₀^∞ e^(−r²) · r dr

The inner integral is elementary: substitute u = r², du = 2r dr, giving ∫₀^∞ (1/2) e^(−u) du = 1/2. So I² = 2π · (1/2) = π, hence I = √π.

The right panel shows the 2D version: a glowing bell on the (x, y) plane whose total volume is exactly π. Slice radially to see where the proof lives. π appears because the 2D bell is rotationally symmetric — and rotation in the plane is the geometric thing π measures.