Problem 53 - Thomas Calculus

Question

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.8 .\) Evaluate the improper integrals in Exercises \(51-54\) as iterated integrals. $$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\left(x^{2}+1\right)\left(y^{2}+1\right)} d x d y $$

Step-by-Step Solution

Answer

The integral evaluates to \( \pi^2 \).

Step 1: Analyze the Integral

We are tasked with evaluating the improper double integral: \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(x^2+1)(y^2+1)} \, dx \, dy \].This is an iterated integral, meaning we will evaluate the inner integral first and then the outer.

Step 2: Evaluate the Inner Integral

First, hold \(y\) constant and evaluate the inner integral with respect to \(x\): \[ \int_{-\infty}^{\infty} \frac{1}{(x^2+1)(y^2+1)} \, dx \].For a specific \(y\), the integral is:\[ \frac{1}{y^2 + 1} \int_{-\infty}^{\infty} \frac{1}{x^2+1} \, dx \].The known result of \(\int_{-\infty}^{\infty} \frac{1}{x^2+1} \, dx\) is \(\pi\) (an arctangent function evaluated at \(x\to\infty\) and \(x\to-\infty\)).Thus, the inner integral evaluates to:\[ \frac{\pi}{y^2 + 1} \].

Step 3: Setup the Outer Integral

Using the result from Step 2, the outer integral becomes:\[ \int_{-\infty}^{\infty} \frac{\pi}{y^2+1} \, dy \].

Step 4: Evaluate the Outer Integral

The outer integral \[ \pi \int_{-\infty}^{\infty} \frac{1}{y^2+1} \, dy \].Again, this is the integral of an arctangent function, just like in Step 2. Its value is \(\pi\).Therefore, the outer integral evaluates to:\[ \pi^2 \].

Step 5: Conclusion: Final Result

The original improper double integral evaluates to: \[ \pi^2 \].

Key Concepts

Double Integrals

Double integrals are an extension of single-variable integrals to functions of two variables. They are used to compute volume under a surface over a two-dimensional area or region. In mathematical terms, a double integral is represented as \[ \int \int f(x, y) \, dA \]where \(f(x, y)\) is a function of \(x\) and \(y\), and \(dA\) represents an infinitesimal area element in the \(xy\)-plane.
Double integrals, especially improper ones, can evaluate complex integrals over infinite regions or when the integrand is improperly behaved. The process often involves splitting the integral into iterated integrals—one for each variable.

• Iterated Integrals: Breaks the double integral into two separate integrations done sequentially.
• Improper Double Integrals: Extends the idea of improper integrals from single-variable calculus, often involving limits.

Double integrals are essential in fields like physics and engineering to calculate mass, area, volumes, and center of mass for objects with varying densities.

Iterated Integrals

In the context of evaluating double integrals, the term iterated integrals refers to performing integration one variable at a time, proceeding in a fixed order. This approach transforms a double integral into two single integrals, one nested within the other. Here is the general setup:
\[ \int_a^b \int_c^d f(x, y) \, dx \, dy \]
This expression indicates that we first integrate \(f(x, y)\) with respect to \(x\) while holding \(y\) constant, followed by integrating with respect to \(y\).

• Inner Integral: The first integral performed, taking one variable out of the equation temporarily.
• Outer Integral: The subsequent integral using the result of the inner integration to integrate over the second variable.

Iterated integrals are particularly useful because they break down complicated calculations into manageable steps, each of which considers only one variable at a time.

Arctangent Function

The arctangent function, denoted by \(\tan^{-1}(x)\) or \(\arctan(x)\), is the inverse of the tangent function. It returns the angle whose tangent is a given number. When used in integrals, it often arises in contexts where the integrand contains terms like \(\frac{1}{x^2 + a^2}\), indicating a trigonometric substitution might be suitable.
One fundamental integral associated with the arctangent function is:\[ \int \frac{1}{x^2 + 1} \, dx = \arctan(x) + C \]
This result is crucial in problems involving improper integrals extending over \(-\infty\) to \(\infty\), such as evaluating:\[ \int_{-\infty}^{\infty} \frac{1}{x^2 + 1} \, dx = \pi \]
In the provided exercise, this property helped simplify the double integral evaluation process.

• Applications: Arctangent functions appear in physics, engineering, and geometry involving angle calculations.
• Properties: It maps real numbers to angles, typically in the range of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

Understanding the behavior and integration of the arctangent function is key to handling problems involving unit circular motion, resonance, and wave analysis.