Question

Prove that \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}\). (Hint: It is not difficult to show that \(I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y\). By a change of variables to polar coordinates prove that \(I^{2}=\pi\).)

Short answer

The given integral \( I = \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \).

Step-by-step solution

Interpret the Hint

The given problem asks us to prove that \( I = \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi} \). The hint suggests proving this by first showing that \( I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy \). This involves converting the double integral into polar coordinates to show that \( I^2 = \pi \).

Set Up the Double Integral

We express \( I^2 \) as a product of two identical integrals: \( I = \int_{-\infty}^{\infty} e^{-x^2} \, dx \), so \( I^2 = \int_{-\infty}^{\infty} e^{-x^2} \, dx \times \int_{-\infty}^{\infty} e^{-y^2} \, dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy \).

Convert to Polar Coordinates

We convert \( (x, y) \) to polar coordinates where \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The expression \( x^2 + y^2 \) becomes \( r^2 \), simplifying the integral. The area element \( dx \, dy \) in Cartesian coordinates is replaced by \( r \, dr \, d\theta \) in polar coordinates, so the integral becomes \( \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r \, dr \, d\theta \).

Evaluate the Integral

First, evaluate the integral with respect to \( r \): \( \int_0^{\infty} e^{-r^2} r \, dr \). Let \( u = r^2 \), then \( du = 2r \, dr \), simplifying to \( \frac{1}{2} \int_0^{\infty} e^{-u} \, du = \frac{1}{2} [-e^{-u}]_0^{\infty} = \frac{1}{2} \).

Integrate Over \( \theta \)

Now integrate over \( \theta \), \( \int_0^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times 2\pi = \pi \).

Solve for \( I \)

Since \( I^2 = \pi \), taking the square root on both sides gives \( I = \sqrt{\pi} \).

Key concepts

Double Integral

A double integral is an essential method in calculus used to evaluate the accumulation of quantities over a two-dimensional area. While a single integral sums up quantities along a line or curve, a double integral does so over an area or surface. Double integrals have the specific form:\[ \int_{a}^{b} \int_{c}^{d} f(x, y) \, dx \, dy \]In this specific mathematical proof to solve the Gaussian Integral, the function in consideration is \(e^{-(x^2 + y^2)}\). The product of two separate integrals \(\int_{-\infty}^{\infty} e^{-x^2} \, dx\) and \(\int_{-\infty}^{\infty} e^{-y^2} \, dy\) leads to a double integral:\[ I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy \]This integral spans the entire plane, capturing the nature of the Gaussian function in two dimensions, allowing us to evaluate it comprehensively by transforming to polar coordinates for simplification.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system in which a point on a plane is determined by an angle and a distance from a reference point. This system is particularly useful for problems with circular symmetry, like the problem in question. By converting Cartesian coordinates \((x, y)\) into polar coordinates \((r, \theta)\), where:

• \( x = r \cos(\theta) \)
• \( y = r \sin(\theta) \)

the expression \(x^2 + y^2\) becomes \(r^2\), thus simplifying the integrand \( e^{-(x^2 + y^2)} \) to \( e^{-r^2} \). In this transformation:

• \( dx \, dy \) in Cartesian becomes \( r \, dr \, d\theta \) in polar coordinates.

This reduction is pivotal because it changes a complex integral in Cartesian coordinates to one that is much simpler to solve in polar form. This is why transforming into polar coordinates is a favored method for tackling Gaussian integrals across the plane.

Integration Techniques

Integration is a fundamental concept in calculus that finds the integral of a function. Various techniques can be used depending on the function's form and the domain of integration. For the task at hand, involving the Gaussian integral, the technique involves a change of variables, specifically transitioning from Cartesian to polar coordinates. In polar coordinates, once the transformation is done, evaluating the integral involves solving:\[ \int_0^{\infty} e^{-r^2} r \, dr \]Here, a substitution is useful:

• Set \( u = r^2 \), then \( du = 2r \, dr \), simplifying the integrand to \( \frac{1}{2} \int_0^{\infty} e^{-u} \, du \).

The integral \( \int e^{-u} \, du \) is directly computable and yields \(-e^{-u} + C\), where evaluating it at the boundary (0 to \(\infty\)) yields \(\frac{1}{2}\). Techniques like substitution streamline the integration process, and this problem exemplifies using a combination of coordinate transformation and substitution for efficient computation.

Mathematical Proof

A mathematical proof is a logical argument that establishes the truth of a statement rigorously. In mathematics, proofs must be valid and follow from previously accepted principles. In this problem, the goal is to prove that the integral of the Gaussian function is equal to the square root of \(\pi\). Here's how this proof progresses:

• Firstly, interpret the problem's hint which guides proving that \( I^2 = \pi \).
• Next, use the concept of a double integral to represent this as \( I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy \).
• Apply a variable transformation to simplify this integral into polar coordinates, making the calculations manageable.
• Once simplified, solve the integral with respect to \( r \) and then integrate over \( \theta \).
• Finally, show that this results in \( \pi \), thus \( I = \sqrt{\pi} \) when the square root is taken on both sides.

This chain of logical steps, supported by integration techniques and coordinate transformations, validates the original statement through proven mathematical principles.