Question

In Exercises 23-26, evaluate the improper iterated integral. $$ \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{x y} d x d y $$

Short answer

The double integral diverges since the inner integral diverges.

Step-by-step solution

Evaluate the Inner Integral

First, we solve the inner integral. We integrate the function \(\frac{1}{xy}\) with respect to x, treating y as a constant. The inner integral as follows: \(\int_{1}^{\infty} \frac{1}{x y} d x\) becomes \(\frac{1}{y} \int_1^\infty \frac{1}{x} d x\). Given the integral of \(\frac{1}{x}\) is \(ln x\), we have \(\frac{1}{y} [ln x ]_1^\infty\). Evaluating limits gives \(\frac{1}{y}(ln \infty - ln 1) = \frac{1}{y} (\infty - 0) = \infty\). Thus, the inner integral diverges.

Evaluate the Outer Integral

Next, we move to the outer integral. However, as the inner integral is divergent, the entire double integral is divergent. There is no need to compute the outer integral since the concept of double integral assumes that the inner integral converges.

Concluding the Solution

As we saw in steps 1 and 2, if one of the integrals in an improper integral diverges, the entire integral is considered to diverge. Thus, the given double integral also diverges.

Key concepts

Double Integral

Understanding the concept of a double integral is crucial when dealing with functions of two variables. A double integral allows us to compute the volume under a surface or the area of a region in the plane.

A double integral, such as \[ \int \int f(x, y)\, dx\, dy \], is typically evaluated by integrating the function f(x, y) with respect to one variable while considering the other variable to be constant, and then integrating the resulting expression with respect to the second variable. This process is known as iterated integration.

It's important to remember that for the double integral to exist, the integrals must converge, which essentially means that they produce a finite area or volume. If any one of the iterated integrals diverges, as seen in the example discussed in the exercise, the entire double integral cannot be evaluated and is thus considered to diverge.

Convergence and Divergence of Integrals

The convergence and divergence of an integral are fundamental concepts in calculus concerning whether an integral yields a finite value (converges) or not (diverges).

For an integral to converge, the function being integrated must approach a finite limit as the variable approaches the integration limit. If the function does not approach a finite limit, the integral will 'run off to infinity', meaning it diverges. This is particularly relevant for improper integrals, where the integration limits include infinity or where the integrand has an infinite discontinuity within the integration interval.

In the provided exercise, the integral \[ \int_{1}^{\infty} \frac{1}{x} dx \] diverges because as x approaches infinity, the function \( \frac{1}{x} \) tends towards zero too slowly to produce a finite area — its integral, the natural logarithm of infinity, is undefined.

Integration Techniques

There are various integration techniques employed to evaluate integrals, each suitable for different kinds of functions and integration scenarios.

Some widely used techniques include:

• Substitution: Used when a function is the composition of a function and its derivative.
• Integration by parts: Applicable when an integral is the product of two functions.
• Partial fraction decomposition: Helpful for integrating rational functions.
• Trigonometric substitution: Used for integrals involving square roots of quadratic expressions.

Improper integrals, such as the example in the exercise, often require a careful approach to determine convergence or divergence. In these cases, it's not just about applying an integration technique but also about understanding the behavior of the function as variables tend towards their limits — an essential step for determining the outcome of the integral.