Problem 69 - 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. $$\int_{1}^{\infty} \int_{e^{-}}^{1} \frac{1}{x^{3} y} d y d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \).

Step 1: Set up the Iterated Integral

We start by writing down the iterated integral as given in the problem. This is the integral \( \int_{1}^{\infty} \int_{e^{-}}^{1} \frac{1}{x^{3} y} \, dy \, dx \). The limits for \( y \) are from \( e^{-} \) to \( 1 \) while the limits for \( x \) are from \( 1 \) to \( \infty \).

Step 2: First Integration with Respect to y

Integrate \( \frac{1}{x^3 y} \) with respect to \( y \). Since the variable \( x \) is treated as a constant during this integration, the integral becomes:\[\int_{e^{-}}^{1} \frac{1}{x^{3} y} \, dy = \frac{1}{x^3} \int_{e^{-}}^{1} \frac{1}{y} \, dy\]The integral \( \int \frac{1}{y} \, dy \) is equal to \( \ln |y| \), so:\[\frac{1}{x^3} \left[ \ln |y| \right]_{e^{-}}^{1} = \frac{1}{x^3} \left ( \ln 1 - \ln e^{-} \right)\]This simplifies to:\[\frac{1}{x^3} \cdot \ln e = \frac{1}{x^3} \]

Step 3: Second Integration with Respect to x

Now integrate \( \frac{1}{x^3} \) with respect to \( x \) from \( 1 \) to \( \infty \):\[\int_{1}^{\infty} \frac{1}{x^3} \, dx = \left[ -\frac{1}{2x^2} \right]_{1}^{\infty}\]Evaluating this limit:- At \( x = \infty \), \(-\frac{1}{2x^2} \to 0\) because the denominator grows very large.- At \( x = 1 \), \(-\frac{1}{2(1)^2} = -\frac{1}{2} \).Thus, the integral evaluates to:\[0 - \left( -\frac{1}{2} \right) = \frac{1}{2}\]

Step 4: Conclusion: Evaluate the Final Result

The value of the improper double integral is \( \frac{1}{2} \) after evaluating both iterated integrals and simplifying the resulting expressions.

Key Concepts

Double Integration

Double integration is a cornerstone of multivariable calculus. It's used to compute the volume under a surface in a three-dimensional space. Imagine a surface floating above a plane - double integration helps measure what's underneath. In this context, it involves finding the integral over a given region by performing integration twice, typically with respect to two variables.

• In our exercise, we're working with the function \( \frac{1}{x^3 y} \).
• We integrate once with respect to \( y \), and then with respect to \( x \).

This two-step integration process allows us to handle complex functions that can’t be solved with a single integration. Double integration is incredibly versatile, handling everything from volumes and areas to more abstract applications in physics and engineering.

Iterated Integrals

Iterated integrals are a specific way to compute double integrals. This method involves breaking the double integral into a sequence of single integrals. By treating one variable as a constant at a time, we're able to simplify the computation significantly.

• Step one is to integrate with respect to the first variable, in this case, \( y \), while considering \( x \) constant.
• After finding the result of this inner integral, we integrate the result with respect to the second variable, \( x \).

Iterated integrals are crucial when dealing with non-rectangular regions or variable limits of integration. By evaluating the integral in stages, we ensure each step is manageable, which simplifies our approach to solving more complex integrals.

Limit Evaluation

Limit evaluation is often necessary when dealing with improper integrals, like in this exercise. Improper integrals occur when either the interval of integration is infinite or the integrand becomes infinite within the interval.
In our problem, the interval for \( x \) stretches to infinity.

• To evaluate this, we calculate the limit as \( x \) approaches infinity for the function \(-\frac{1}{2x^2}\).
• As \( x \) gets larger, the term diminishes towards zero.

This limiting process helps us make sense of infinite behaviors and ensures that our integrals have meaningful, finite values. Ultimately, limit evaluation provides the tools to tackle improper integrals confidently, leading to precise results like \( \frac{1}{2} \) in our solution.