Question

In Exercises 23-26, evaluate the improper iterated integral. $$ \int_{1}^{\infty} \int_{0}^{1 / x} y d y d x $$

Short answer

The value of the improper iterated integral is \( -1 / 2 \).

Step-by-step solution

Perform Inner Integral

First, take the inner integral with respect to \( y \). This integral is a simple power rule problem, so apply the power rule, which states that the integral of \( x^n \) is \( (1/(n+1)) * x^{n+1} \), where \( n \) is the exponent and \( x \) is the base. For this integral, we have \( n = 1 \), so it becomes \( (1/(1+1)) * y^{1+1} = y^2 / 2 \). The integral, thus, yields \( y^2 / 2 \) evaluated from \( y = 0 \) to \( y = 1 / x \).

Evaluate the Definite Integral

Next, substitute \( y = 1 / x \) and \( y = 0 \) into \( y^2 / 2 \) and subtract the two results to evaluate the definite integral. Therefore, we get \( [(1 / x)^2 / 2] - [0] = 1 / 2x^2 \).

Perform Outer Integral

Now, we perform the outer integral with respect to \( x \), which is from 1 to infinity of \( 1 / 2x^2 \). Again using the power rule, the integral of \( x^n \) is \( (1/(n+1)) * x^{n+1} \), so \( - 1 / 2x \) evaluated from 1 to infinity.

Evaluate the Definite Integral

Finally, evaluate the result of the outer integral by substituting \( x = \infty \) and \( x = 1 \) into \( - 1 / 2x \) and subtracting the two results. We obtain \( [- 1 / 2*1] - [0] = -1 / 2 \). Note that the generally infinite term vanishes because of the negative sign in the equation, turning \( 1 / \infty \) to 0.