Problem 83 - University Calculus: Early Transcendentals

Question

Unbounded region Prove that $$\begin{aligned} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y &=\lim _{b \rightarrow \infty} \int_{-b}^{b} \int_{-b}^{b} e^{-x^{2}-y^{2}} d x d y \\ &=4\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)^{2} \end{aligned}$$

Step-by-Step Solution

Answer

The double integral evaluates to \(\pi\), matching the product of the squared single integral and a factor of 4.

Step 1: Understand the Problem

The problem requires us to evaluate a double integral over an infinite region and prove its equivalence to a square of a simpler definite integral. The task is to show that integrating the function \(e^{-x^2-y^2}\) over the entire xy-plane results in the expression \(4\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)^{2}\).

Step 2: Convert to Polar Coordinates

Since \(e^{-x^2-y^2}\) is radially symmetric, it is advantageous to compute the integral in polar coordinates. Let \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), where \(r^2 = x^2 + y^2\). The infinitesimal area element \(dx \ dy\) becomes \(r \ dr \ d\theta\). The integral then becomes: \[ \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \, dr \, d\theta \]

Step 3: Simplify the Integral

Separate the integral into the product of two integrals: \[ \int_{0}^{2\pi} d\theta \cdot \int_{0}^{\infty} e^{-r^2} r \, dr \].The first integral evaluates to \(2\pi\) since it's the complete range over \(\theta\). The second integral, \(\int_{0}^{\infty} e^{-r^2} r \, dr\), requires substitution or recognition as a standard Gaussian integral.

Step 4: Evaluate the Radial Integral

Perform the substitution \( u = r^2 \), which gives \( du = 2r \, dr \), or \( r \, dr = \frac{1}{2} du \). The integral becomes:\[ \frac{1}{2} \int_{0}^{\infty} e^{-u} \, du \] which evaluates to \(\frac{1}{2}\).

Step 5: Conclude the Polar Integral

Multiply the results of the two integrals: \[ 2\pi \times \frac{1}{2} = \pi \].This completes the evaluation of the original double integral as \(\pi\).

Step 6: Relate to Single Integral and Complete the Proof

For the single integral \(\int_{0}^{\infty} e^{-x^2} dx\), its square results in the Gaussian integral over all space:\[ \left(\int_{0}^{\infty} e^{-x^2} dx\right)^2 = \frac{\pi}{4} \] Thus, multiplying out gives:\[ 4 \times \frac{\pi}{4} = \pi \]as required. Therefore, we have shown:\[ \int_{-\infty}^{\ ext{infty}} \int_{-\ ext{\infty}}^{\infty} e^{-x^2 - y^2} dx \ dy = 4 \left( \int_{0}^{\infty} e^{-x^2} dx \right)^2 \].

Key Concepts

Polar Coordinates

The clever conversion to polar coordinates is a game-changer. When dealing with functions involving the sum of squares like \( e^{-x^2-y^2} \), polar coordinates make computations simpler.
Consider changing from Cartesian to polar coordinates with these transformations: \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). The relationship \( r^2 = x^2 + y^2 \) highlights the symmetry which is key for simplification. The area element \( dx \, dy \) morphs into \( r \, dr \, d\theta \).
This transformation capitalizes on the circular symmetry of the original function \( e^{-x^2-y^2} \), condensing the double integral into a much more manageable form. By switching to a radial coordinate \( r \) and angular coordinate \( \theta \), the problem becomes more intuitive, especially within circular or symmetric regions, emphasizing the beauty of polar integration over such infinite domains.

Gaussian Integrals

Gaussian integrals feature a quintessential part of calculus with their elegant closed-form solutions. The Gaussian function \( e^{-x^2} \) may look complex, but when integrated, it miraculously produces \( \sqrt{\pi} \).
In the original problem, the key realization is that we can express a two-dimensional problem involving \( e^{-x^2-y^2} \) in terms of the simpler, one-dimensional integral \( \int_{0}^{\infty} e^{-x^2} \, dx \). When squared and multiplied by four, this results in \( \pi \).
The trick lies in recognizing this integral as a standard Gaussian integral, revealing an unexpected connection between these integrals and the geometry of circles as it relates to \( \pi \). This highlights one of the most satisfying aspects of calculus: encountering solutions with inherent symmetries and intrinsic beauty, where simple yet profound ideas about area and function overlap to create unification.

Infinite Regions

The concept of infinite regions in integral calculus often seems daunting, but it becomes manageable with the right techniques. Here, you encounter the infinite plane in the limits \( -\infty \) to \( \infty \) for both \( x \) and \( y \).
This interval captures the whole of the xy-plane, making the integration cover entirely an infinite spatial extent. The sum over these large regions can sometimes be remarkably finite, thanks to the rapid decay of Gaussian-type functions.
In this problem, using polar coordinates and acknowledging the symmetry of the Gaussian facilitates handling these infinite areas. They are typically approached by taking limits as the boundaries of a region grow towards infinity. This sometimes includes considering portions of the plane, such as quadrants, and scaling results appropriately to arrive at full-space solutions to obtain the correct finite result.