Question

Existence of improper double integral For what values of \(m\) and \(n\) does the integral \(\int_{1}^{\infty} \int_{0}^{1 / x} \frac{y^{m}}{x^{n}} d y d x\) have a finite value?

Short answer

In order for the improper double integral to have a finite value, the values of m and n must satisfy the condition (m+n) > -1.

Step-by-step solution

Evaluate the integral with respect to y

Let's start by evaluating the integral with respect to \(y\). The integral we have is: $$ \int_{1}^{\infty} \int_{0}^{1 / x} \frac{y^{m}}{x^{n}} dy dx $$ Now, let's focus on the inner integral: $$ \int_{0}^{1 / x} y^{m} dy $$ Using the power rule for integration, we get: $$ \frac{y^{m+1}}{m+1} \Big|_0^{1/x} $$ Plugging in the limits of integration, we get: $$ \frac{(1/x)^{m+1}}{m+1} - \frac{0^{m+1}}{m+1} = \frac{1}{(m+1) x^{m+1}} $$ Now we can rewrite the original double integral as: $$ \int_{1}^{\infty} \frac{1}{x^{n}} \left(\frac{1}{(m+1) x^{m+1}}\right) dx $$

Combine terms and evaluate the integral with respect to x

Now we will combine the terms in the integral and evaluate the integral with respect to \(x\). Combining the terms, we get: $$ \frac{1}{(m+1)} \int_{1}^{\infty} \frac{1}{x^{m+n+1}} dx $$ This integral can be evaluated using the power rule for integration, and we will get: $$ \frac{1}{(m+1)} \left(-\frac{1}{m+n} x^{-(m+n)}\right) \Big|_1^{\infty} $$

Evaluate the limits of integration and determine the conditions for convergence

Next, we will plug in the limits of integration and determine the conditions for convergence. First, let's plug in the upper limit of integration: $$ \frac{-1}{(m+1)(m+n)} \frac{1}{x^{m+n}} \Big|_{x=\infty} = 0 $$ The term goes to zero as \(x\) approaches infinity. Now let's plug in the lower limit of integration: $$ \frac{-1}{(m+1)(m+n)} \frac{1}{x^{m+n}} \Big|_{x=1} = -\frac{1}{(m+1)(m+n)} $$ For the double integral to converge to a finite value, the sum of the resulting expression must be finite. Therefore, the condition for convergence is that: $$ (m+n) > -1 $$ In conclusion, for the improper double integral to have a finite value, the values of \(m\) and \(n\) must satisfy \((m+n) > -1\).

Key concepts

Convergence Conditions

In mathematics and calculus, convergence conditions are essential when dealing with improper integrals, such as the one presented in our exercise. An improper integral often includes an infinite limit of integration or an integrand that approaches infinity within the limits of integration. To assess whether such an integral converges (i.e., it has a finite value), we must satisfy specific conditions.

For the integral in question \[\int_{1}^{\infty} \int_{0}^{1 / x} \frac{y^{m}}{x^{n}} d y d x,\] the convergence condition requires that the sum of the exponents in the integrand, from both variables \(m\) and \(n\), must be analyzed.

Through the evaluation steps, it becomes evident that the integral will converge if

• the resulting expression after integration is finite at its limits,
• specifically, if \((m+n) > -1\).

This condition ensures that the values achieved in the integration process progressively approach a definite number and thus composes an integral value.

Power Rule for Integration

Integrating functions is a fundamental aspect of calculus, and the power rule for integration is one of the most straightforward techniques. This rule is specifically applicable to functions of the form \(x^n\), where \(n\) is any real number except \(-1\).

The power rule states that:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C,\]where \(C\) represents the constant of integration.

It is particularly useful in this exercise when dealing with the inner integral: \[\int_{0}^{1 / x} y^{m} dy.\]Utilizing the power rule allows for simplification and solution of the integrand to become

• \(\frac{y^{m+1}}{m+1} \Big|_0^{1/x} = \frac{1}{(m+1)x^{m+1}}\).

This straightforward application demonstrates how the power rule helps in transforming complex integrals into simpler expressions easy to evaluate.

Double Integration

Double integration involves integrating a function over a two-dimensional area. In the exercise provided, the function being integrated is \(\frac{y^m}{x^n}\), and the region of integration is determined by the limits for both \(y\) and \(x\).

The process of double integration typically proceeds in sequential steps using nested integrals. First, we integrate with respect to one variable, then use that result to integrate with respect to the second variable. This approach was directly applied in the example:

• First, the integral with respect to \(y\) is resolved, yielding a function dependent solely on \(x\).
• Second, this function is then the subject of the subsequent integral with respect to \(x\).

The end result provides a solution over the specified area of integration.

Double integrals are integral to understanding volume under surfaces or, more generally, areas in higher dimensions, thereby extending the fundamental principles of single-variable calculus.