Problem 83 - Thomas Calculus

Question

Unbounded region Prove that \begin{equation}\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\end{equation} \begin{equation} \quad\quad\quad\quad\quad\quad=4\left(\int_{0}^{\infty} e^{-x^{2}} d x\right)^{2} \end{equation}

Step-by-Step Solution

Answer

The double integral equals \(\pi\), confirming \(4 \left( \int_{0}^{\infty} e^{-x^2} dx \right)^2\).

Step 1: Understand the Problem

We need to prove that the double integral of the function \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} dx dy \]over the entire plane is equal to \[ 4 \left( \int_{0}^{\infty} e^{-x^{2}} dx \right)^2. \]This involves proving that as the limit goes to infinity, the bounded integral equals a certain expression with a squared single-variable integral.

Step 2: Change of Order of Integration

The problem hints at using symmetry and the ability to change integration orders on decaying exponential functions. Recognize that since \(e^{-x^2-y^2}\) separates into a product of individual functions in \(x\) and \(y\), this allows simplifications using polar coordinates.

Step 3: Convert to Polar Coordinates

Transform the double integral into polar coordinates:\[ x = r \cos \theta, \quad y = r \sin \theta \]with \( dx \, dy = r \, dr \, d\theta \). The integral becomes\[ \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \, dr \, d\theta. \]

Step 4: Evaluate the Inner Integral

Perform the inner integral with respect to \(r\):\[ \int_{0}^{\infty} e^{-r^2} r \, dr. \]Let \(u = r^2\), then \(du = 2r \, dr\), so \(r \, dr = \frac{1}{2} du\). The integral becomes:\[ \frac{1}{2} \int_{0}^{\infty} e^{-u} du = \frac{1}{2}. \]

Step 5: Integrate with Respect to Theta

Now, integrate with respect to \(\theta\):\[ \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \cdot 2\pi. \]

Step 6: Simplify the Result

This simplification gives us the total integral value \(\pi\). Verify if this setup matches the structure given in the problem statement about single integral squares.

Step 7: Verify Relation to Square of Single Integral

Recall that \( (\int_{0}^{\infty} e^{-x^2} \, dx) \) evaluates to \( \frac{\sqrt{\pi}}{2} \). Then:\[ \left( \int_{0}^{\infty} e^{-x^2} \, dx \right)^2 = \left( \frac{\sqrt{\pi}}{2} \right)^2 = \frac{\pi}{4}. \]Multiplying by 4 gives \(\pi\), confirming that the left side matches the intended limit value.

Key Concepts

Polar Coordinates

When tackling a double integral problem over an infinite region, converting to polar coordinates can be incredibly useful. This is because polar coordinates are often a natural fit for circular or radial symmetry inherent in equations involving variables like \(x^2 + y^2\). The conversion formulas from rectangular to polar coordinates are \(x = r \cos \theta\) and \(y = r \sin \theta\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle from the positive x-axis.

When converting a double integral from Cartesian coordinates (\(x, y\)) to polar coordinates (\(r, \theta\)), the differentials change as well, becoming \(dx \, dy = r \, dr \, d\theta\). Incorporating this into the integral allows it to account for the scaling by the factor of \(r\), which accounts for the area element difference between the two coordinate systems.

• Polar coordinates simplify integration limits for radial constraints.
• The variable \(r\) always remains non-negative, simplifying calculations.
• Angular integration often runs from \(0\) to \(2\pi\) for full circles.

These properties are crucial when dealing with functions defined over circular regions or infinite planes.

Exponential Function

The exponential function \(e^{-x^2-y^2}\) is key to understanding many problems in integral calculus, especially when they arise in the context of infinite domains. Each variable, \(x\) and \(y\), independently decays exponentially. This property allows the function to factor into the product of two simpler one-dimensional functions, which significantly simplifies calculations as you shift into polar coordinates.

The function \(e^{-r^2}\) retains its essential behavior under transformation to polar, emphasizing its core advantage:

• Decays rapidly to zero, which is useful over infinite domains.
• Preserves symmetry, allowing straightforward application of symmetry-based theorems.
• Integrals over Gaussian functions often lead to results that involve terms like \(\sqrt{\pi}\), due to the properties of the Gaussian (bell curve) function.

Recognizing the separability of the exponential function in rectangular coordinates enables more efficient evaluation in polar settings. This separation is essential for analytically solving many otherwise complex integrals.

Change of Order of Integration

The change of order of integration is an integral calculus technique that can simplify double integrals by transforming the limits and sequence of integration. It relies on Fubini's theorem, which allows reversing the order when both iterated integrals converge. The technique is particularly useful when one order of integration is more challenging than the other.

When dealing with functions that express symmetry or decay, like \(e^{-x^2-y^2}\), the ingenuity lies in recognizing aspects like symmetry in order to reorganize the limits into forms that are solvable. You might often encounter this in problems involving rotational symmetry or when using alternative coordinate systems like polar coordinates, where the regions of integration become more natural rectangles.

• Look for symmetry in the integrand to simplify problem-solving.
• Evaluate feasibility and convenience of each order of integration.
• Utilize change of variables to make integral limits significantly more manageable.

Understanding when and how to change the order of integration is a vital skill, especially when dealing with complex regions or unbounded domains.

Integral Calculus

Integral calculus is the study of integrals and is a fundamental part of mathematical analysis dealing with areas, volumes, and central concepts like accumulation and net change. In the provided exercise, integral calculus helps evaluate the double integrals over an infinite plane. This involves both recognizing the form of the integral and applying techniques like substitution and integration by parts.

Evaluating integrals with respect to one variable over a bounded interval and then extending them to an unbounded interval (as with limits) is common in proving identities and understanding components of infinite integrals. Calculating such integrals requires knowledge not only of how to operate with limits and bounds but also of leveraging fundamental theorems of calculus to transition from infinite double integrals to expressions in terms of one-dimensional integrals.

• Utilize special integrals and known results in difficult computations.
• Apply numerical methods, if necessary, for evaluating definite integrals.
• Regularly employ symmetry arguments to reduce complexity.

Integral calculus is the cornerstone of mathematical problem-solving and modeling, offering both exact and approximate solutions to diverse, often continuous, problems.