Problem 18 - Calculus and its applications

Question

Determine whether each improper integral is convergent or divergent, and find its value if it is convergent. $$ \int_{1}^{\infty} \ln x d x $$

Step-by-Step Solution

Answer

The integral is divergent.

Step 1: Define the Integral as a Limit

To handle the integral from 1 to infinity, we express it as a limit: \[ \int_{1}^{ ext{c}} \ln x \, dx \] as \( c \to \infty \). We will evaluate the integral first with an upper limit \( c \) and then later take the limit as \( c \to \infty \).

Step 2: Find the Antiderivative

Compute the antiderivative of the integrand \( \ln x \). We can use integration by parts where we set \( u = \ln x \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \). Applying integration by parts formula, \[ \int \ln x \, dx = x \ln x - \int x \frac{1}{x} \, dx = x \ln x - x + C \]. So, \( F(x) = x \ln x - x + C \) is the antiderivative.

Step 3: Evaluate the Integral from 1 to c

Substitute \( F(x) = x \ln x - x \) to evaluate the integral from 1 to \( c \): \[ \int_{1}^{c} \ln x \, dx = \left[ x \ln x - x \right]_{1}^{c} = \left( c \ln c - c \right) - \left( 1 \ln 1 - 1 \right) = c \ln c - c + 1 \].

Step 4: Take the Limit as c Approaches Infinity

Now, determine whether the integral is convergent or divergent by taking the limit \( c \to \infty \): \[ \lim_{c \to \infty} (c \ln c - c + 1) \]. As \( c \to \infty \), \( c \ln c \) grows faster than \( c \), leading to \( \lim_{c \to \infty} (c \ln c - c + 1) = \infty \).

Step 5: Determine Convergence or Divergence

Since the limit \( \lim_{c \to \infty} (c \ln c - c + 1) \) equals infinity, the integral \int_{1}^{\infty} \ln x \, dx is divergent rather than convergent.

Key Concepts

Limit of Integration

Improper integrals often involve limits in one or both of their bounds. Instead of having fixed limits like 0 or 1, indefinite integrals can extend to infinity. To compute these, we define the integral as a limit. For example, in the integral \[ \int_{1}^{\infty} \ln x \, dx \]we express it as:\[ \lim_{c \to \infty} \int_{1}^{c} \ln x \, dx \]This approach allows us to evaluate the integral within a finite interval first and then consider the behavior as the upper limit approaches infinity. Since we cannot directly plug infinity into our calculations, the concept of the limit is essential to handle integrals extending indefinitely. By evaluating these limits, we can determine if the entire improper integral converges to a finite value or diverges to infinity.

Integration by Parts

The process of finding the antiderivative of a function like \( \ln x \) often requires a strategic technique known as integration by parts. This technique is based on the product rule for differentiation and is used to transform the integral of a product of functions into an easier form.To apply integration by parts, we first identify parts of our function:

• Choose \( u \) and \( dv \): For \( \int \ln x \, dx \), we set \( u = \ln x \) and \( dv = dx \).
• Compute \( du \) and \( v \): Differentiate and integrate to find \( du = \frac{1}{x} dx \) and \( v = x \).

Next, apply the integration by parts formula:\[ \int u \, dv = uv - \int v \, du \] For \( \ln x \), this results in:\[ x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - x + C \] The antiderivative, found through this method, helps in evaluating definite integrals, especially those with infinite bounds.

Antiderivative

An antiderivative is a function that reverses the process of differentiation, recovering the original function from its derivative. It's crucial for evaluating integrals because once we have the antiderivative, we can calculate the integral over any interval by the Fundamental Theorem of Calculus.For instance, in the exercise, we find the antiderivative of \( \ln x \). By using integration by parts, the antiderivative is determined to be:\[ F(x) = x \ln x - x + C \]Here, \( C \) is the constant of integration, which is not required when calculating definite integrals as it cancels out. Applying this antiderivative in definite integrations facilitates finding the difference between the values at the upper and lower limits, forming the method’s backbone for calculating areas under curves.

Convergence and Divergence

Convergence and divergence are crucial concepts when dealing with improper integrals. They help us understand whether an integral results in a finite number (convergent) or tends to infinity (divergent).When evaluating\[ \lim_{c \to \infty} (c \ln c - c + 1) \]we observe that as \( c \) increases, the expression grows without bound due to the term \( c \ln c \). Thus, the integral does not settle at a finite value, which indicates that it diverges.Determining convergence or divergence is vital in calculus for understanding the behavior of functions over infinite intervals. While convergent integrals approximate finite quantities, divergent ones highlight growing, uncontrollable behavior, adding further insight into the nature of the function considered.