Problem 71 - University Calculus: Early Transcendentals

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 \(8.7 .\) Evaluate the improper integrals 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 value of the integral is \(\pi^2\).

Step 1: Set Up the Iterated Integral

The given double integral is: \[ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{(x^2+1)(y^2+1)} \, dx \, dy \]First, set up the integral by considering one variable at a time. Let's first integrate with respect to \(x\). This transforms the integral into:\[ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} \frac{1}{(x^2+1)(y^2+1)} \, dx \right) dy \]

Step 2: Compute the Inner Integral with respect to x

Focus on computing the inner integral:\[ \int_{-\infty}^{\infty} \frac{1}{x^2+1} \, dx \]The function \(\frac{1}{x^2+1}\) is the integrand of the arctangent function, whose integral is known to be:\[ \int \frac{1}{x^2+1} \, dx = \arctan(x) + C \]Thus, the limits of integration from \(-\infty\) to \(\infty\) give us:\[ \lim_{A \to \infty} \int_{-A}^{A} \frac{1}{x^2+1} \, dx = \lim_{A \to \infty} \left[ \arctan(x) \right]_{-A}^{A} = \lim_{A \to \infty} [\arctan(A) - \arctan(-A)] \]Since \(\arctan(A)\) approaches \(\frac{\pi}{2}\) and \(\arctan(-A)\) approaches \(-\frac{\pi}{2}\), the result is \(\pi\).

Step 3: Compute the Outer Integral with respect to y

Now, use the result from Step 2 to evaluate the outer integral:\[ \int_{-\infty}^{\infty} \frac{\pi}{y^2+1} \, dy \]This integral is similar to the inner integral and is also an improper integral of the arctangent:\[ \int \frac{1}{y^2+1} \, dy = \arctan(y) + C \]Evaluate the limits from \(-\infty\) to \(\infty\):\[ \lim_{B \to \infty} \int_{-B}^{B} \frac{\pi}{y^2+1} \, dy = \pi \cdot \lim_{B \to \infty} \left[ \arctan(y) \right]_{-B}^{B} = \pi \cdot \lim_{B \to \infty} [\arctan(B) - \arctan(-B)] \]This simplifies to \(\pi \times \pi = \pi^2\), since \(\arctan(B)\) and \(\arctan(-B)\) approach \(\frac{\pi}{2}\) and \(-\frac{\pi}{2}\) respectively.

Key Concepts

Double Integrals

Double integrals are an extension of single-variable calculus into higher dimensions. They are similar to double sums but applied to continuous functions over two dimensions. Double integrals allow us to compute the volume under a surface over a specific region. To solve a double integral, we integrate one variable first, holding the others constant. This is done within specified bounds, and then integrate the result with respect to the second variable. In our example, we first integrate with respect to \(x\) and use the result to integrate with respect to \(y\). By decomposing the problem, we simplify the process of computing integrals over complex regions.

• A double integral is written as \(\int\int f(x, y) \, dx \, dy\).
• Integration can often switch between \(dx\) and \(dy\) if other conditions allow.
• Care should be taken with bounds when variables are vastly different dimensions like \(-\infty \,to\, +\infty\).

Iterated Integrals

Iterated integrals transform multi-variable integration into a sequence of single-variable integrations. This technique simplifies the computation of multiple integrals by addressing each dimension sequentially.In the problem at hand, this method is employed by first integrating with respect to \(x\), then using that result to perform integration concerning \(y\). Iterated integrals are especially practical for improper integrals, where we handle infinite intervals via limits.

• Begin with one variable, keeping the other constant, then reverse the integration process.
• Useful for simplifying multidimensional problems.
• Reduces complexity by handling one variable dimension at a time.

Arctangent Function

The arctangent function, denoted as \( ext{arctan}(x)\), arises naturally when integrating functions with the form \(\frac{1}{x^2 + 1}\). It is the inverse of the tangent function and returns an angle whose tangent is the given number.In calculus, the definite integral of \(\frac{1}{x^2 + 1}\) between \(-\infty\) and \(\infty\) results in \(\pi\), because the arctangent approaches \(\frac{\pi}{2}\) and \(-\frac{\pi}{2}\) at endpoints.

• Important identities: \(\frac{d}{dx} [\arctan(x)] = \frac{1}{x^2+1}\).
• The integral of \(\frac{1}{x^2+1}\) turns into calculating arctan limits.
• Key in evaluating improper integrals with trigonometric expressions.