∫ e^(-x^2) dx from -∞ to ∞

First, call the integral I:

I = ∫ e^(-x^2) dx from -∞ to ∞

Square I, which is to say, multiply it by itself (with y instead of x):

I^2 = [∫ e^(-x^2) dx from -∞ to ∞][∫ e^(-y^2) dy from -∞ to ∞]

Using Fubini's theorem, we can combine these integrals into a double integral, and multiply the integrands:

I^2 = [∫ ∫ e^(-x^2 - y^2) dx dy] from [-∞ to ∞] and [-∞ to ∞]

Now, we switch to polar coordinates, so:

dx dy = r dr dθ
x = r cos(θ)
y = r sin(θ)

We can describe the new limits as follows:

0 ≤ r ≤ ∞
0 ≤ θ ≤ 2π

I^2 = [∫ ∫ e^(-r^2(sin^2(θ) + cos^2(θ))) r dr dθ] from [0 to ∞] and [0 to 2π]
I^2 = [∫ ∫ e^(-r^2) r dr dθ] from [0 to ∞] and [0 to 2π]

∫ e^(-r^2) r dr from [0 to ∞]

u = -r^2
du = -2r dr
dr = -1/(2r) du

(-1/2) ∫ e^(u) du from [0 to ∞]
(-1/2) [e^u] from [0 to ∞]
(-1/2) [e^(-r^2)] from [0 to ∞]

The upper limit will go to zero, and the lower limit will go to -(-1/2) = 1/2.

∫ 1/2 dθ from [0 to 2π]
1/2 [θ] from [0 to 2π]
= π

I^2 = π
I = sqrt(π)