Evaluate the improper integral from 0 to infinity of e^(-x^2) dx.
To evaluate the improper integral, we can use a trick involving polar coordinates. First, consider the following integral:

I = ∫[0,∞] e^(-x^2) dx

Now square this integral and switch to polar coordinates:

I^2 = (∫[0,∞] e^(-x^2) dx)^2
    = ∬[0,∞]x[0,∞]y e^(-(x^2+y^2)) dx dy

Switching to polar coordinates, let x = r cos(θ) and y = r sin(θ). Then,

dx dy = r dr dθ

and 0 ≤ θ ≤ π/2 since x and y are both non-negative.

So our integral becomes:

I^2 = ∬[0,π/2][0,∞] e^(-r^2) * r dr dθ
    = (1/2) ∫[0,π/2] dθ ∫[0,∞] re^(-r^2) dr

Now, make the substitution u = r^2 and du = 2r dr:

I^2 = (1/4) ∫[0,π/2] dθ ∫[0,∞] e^(-u) du
    = (1/4) [θ]_0^(π/2) [-e^(-u)]_0^∞
    = (1/4) (π/2) (1 - 0)
    = π/8

Taking the square root of both sides, we get:

I = √(π/8)

Thus, the value of the improper integral is:

∫[0,∞] e^(-x^2) dx = √(π/8).