Problem 65 - Thomas Calculus

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 integral evaluates to \(\pi\), confirming the expression.

Step 1: Define the Double Integral

We start by considering the double integral over the entire \(xy\)-plane of the function \(e^{-x^2-y^2}\): \[ I = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} \, dx \, dy.\] This represents the area under the curve \(e^{-x^2-y^2}\) over the entire plane.

Step 2: Switch to Polar Coordinates

To simplify the integral, change from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). The conversion is \(x = r\cos(\theta)\), \(y = r\sin(\theta)\), and the Jacobian of this transformation is \(r\). Thus, the integral becomes: \[ I = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} \, r \, dr \, d\theta. \]

Step 3: Integrate Over \(r\)

Focus on the inner integral with respect to \(r\): \[ \int_{0}^{\infty} e^{-r^2} \, r \, dr. \] Use the substitution \(u = r^2\), then \(du = 2r \, dr\), which gives: \[ \frac{1}{2} \int_{0}^{\infty} e^{-u} \, du = \frac{1}{2} \left[ -e^{-u} \right]_{0}^{\infty} = \frac{1}{2}. \] Thus, the result of the \(r\) integration is \(\frac{1}{2}\).

Step 4: Integrate Over \(\theta\)

Now, integrate the constant result \(\frac{1}{2}\) with respect to \(\theta\): \[ \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \, \left[ \theta \right]_{0}^{2\pi} = \frac{1}{2} \times 2\pi = \pi. \]

Step 5: Relate to One-Dimensional Integral

Recall that: \[ \left( \int_{0}^{\infty} e^{-x^2} \, dx \right)^2 = \left( \frac{\sqrt{\pi}}{2} \right)^2 = \frac{\pi}{4}. \] Therefore, \[ 4 \left( \int_{0}^{\infty} e^{-x^2} \, dx \right)^2 = \pi. \] This confirms the original expression: \[ I = 4 \left( \int_{0}^{\infty} e^{-x^2} \, dx \right)^2 = \pi. \]

Key Concepts

Polar Coordinates

When dealing with double integrals over the entire plane, switching to polar coordinates can simplify the math significantly. A double integral in Cartesian coordinates

• requires both variables \( x \) and \( y \).
• In polar coordinates, these become radial distance \( r \) and angle \( \theta \).

The conversion equations are:

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

Using these, the differential \( dA = dx \, dy \) simplifies with the Jacobian conversion factor to \( r \, dr \, d\theta \).This means our integral is transformed from\[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} \, dx \, dy\]to a much simpler form\[\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} \cdot r \, dr \, d\theta.\]Polar coordinates are particularly advantageous when dealing with circular symmetry or functions like \( e^{-x^2-y^2} \), which depend only on \( r^2 = x^2 + y^2 \). They help reduce complexity by aligning with the shape of the function's symmetry.

Change of Variables

The change of variables technique in integrals often involves substituting one set of variables with another to simplify the integration process. When we encounter a double integral like:\[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} \, dx \, dy,\]this transformation is key.In this exercise, we moved from Cartesian to polar coordinates. The equation

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

allowed us to convert the original variables into \( r \) and \( \theta \). This made it easier to perform the integration because it exploited the symmetry of the function, simplifying the limits and the integrand itself.With the Jacobian, a determinant that scales the new variables appropriately, we applied the change of variables correctly: the \( dx \, dy \) became \( r \, dr \, d\theta \) because the area element is scaled by \( r \).By focusing on transformation, this approach allows us to rewrite our integral in a way that is often more straightforward and thus easier to solve.

Improper Integrals

An improper integral is a type of integral where either the function is unbounded or the region of integration is infinite. In this problem, we handle both types:

• The function \( e^{-x^2-y^2} \) tends toward zero but covers an infinite plane.
• The integral limits extend to infinity.

This specific integral appears as:\[\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2} \, dx \, dy,\]which poses the challenge of managing infinite bounds. By switching to polar coordinates, we effectively simplified this to an integral with finite limits on the angle \( \theta \), from \( 0 \) to \( 2\pi \), while the radial part \( r \) remains infinite. Handling \( r \, dr \) from \( 0 \) to \( \infty \) required substitution to tackle its improper nature, dividing the variable, and ultimately applying known conclusions from such standard integrals (like the Gaussian integral) give us: \[\frac{1}{2} \int_{0}^{\infty} e^{-u} \, du = \frac{1}{2}.\]Thus, evaluating improper integrals allows us to conquer problems with infinite limits in a structured and mathematically sound way.