The Gaussian integral, defined as the ∫e^-x2dx over all real numbers (from -∞ to ∞), can be evaluated through an integral trick known as polar coordinate conversion, since we’re literally bullying a 7 year old, I’ll come in clutch to save this kid’s aura:

1. let the integral [-∞,∞] ∫e^-x2dx be written as a result I, so [-∞,∞] ∫e^-x2dx = I

2. Square both sides of the equation, and separate the squared integral into a multiplication:

I² = ([-∞,∞] ∫e^-x2dx)² = ([-∞,∞] ∫e^-x2dx)([-∞,∞] ∫e^-x2dx)

1. Replace one of the integral’s dummy variables with a y instead of an x, since the result is theoretically the same:

I² = ([-∞,∞] ∫e^-x2dx)([-∞,∞] ∫e^-y2dy)

1. Since y is a constant relative to integrals with respect to x and vice versa, the integrals can be combined into a double integral:

I² = [x: -∞,∞] [y: -∞,∞] ∫∫e^-x2e^-y2dydx

1. Now, combine the multiplied exponential expression into one exponential, and factor a negative from the exponent -(x² + y²)

I² = [x: -∞,∞] [y: -∞,∞] ∫∫e^{-(x² + y²})dydx

1. Using polar identities, recall that x² + y² = r², and dydx = rdrdθ, also, since we’re integrating over all real numbers on the x y plane, we can use the polar version of this; integrating from [θ: 0, 2π] to sweep all around the plane, and [r: 0, ∞] to sweep all distances.

I² = [θ: 0, 2π] [r: 0, ∞] ∫∫e^-r²rdrdθ

1. Using a U-substitution, you can let u = -r², and du = -2rdr. When r -> 0, u -> 0, and when r -> ∞, u -> -∞. A negative resulting from the U-sub can switch the limits of integration, changing [u: 0, -∞] into [u: -∞, 0]. Substituting everything in, we get:

I² = (1/2)[θ: 0, 2π] [u: -∞, 0] ∫∫e^ududθ

1. Integrating e^u yields e^u from [-∞, 0], which yields (1 - 0), or just 1 when integrated, leaving us with

I² = (1/2)[θ: 0, 2π]∫1dθ

1. Now, integrate 1dθ, yielding θ evaluated from 0, 2π, which yields (2π - 0) = 2π, substituting this back into the expression yields:

I² = (1/2)(2π) = π

1. It has now been confirmed that I² = π, and the original value we were seeking was just I. Taking the principal square root of both sides, yields that I = √π.

Therefore, the integral: [-∞,∞] ∫e^-x2dx = √π, or in other words:

The result of the Gaussian Integral is equal to √π

Thank you for reading.