Question

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.\(\int_{1}^{\infty} \frac{\ln x}{x} d x\)

Short answer

The integral diverges.

Step-by-step solution

Identify the Type of Improper Integral

The given integral \( \int_{1}^{\infty} \frac{\ln x}{x} \, dx \) is an improper integral because the upper limit of integration is infinite.

Examine the Integrand

The function \( f(x) = \frac{\ln x}{x} \) is continuous for \( x \geq 1 \), so we only have to consider the behavior as \( x \to \infty \).

Apply the Limit Definition of Convergence

Convert the improper integral to a limit: \[\int_{1}^{\infty} \frac{\ln x}{x} \, dx = \lim_{b \to \infty} \int_{1}^{b} \frac{\ln x}{x} \, dx.\]

Compute the Indefinite Integral

To find \( \int \frac{\ln x}{x} \, dx \), use substitution. Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). Thus:\[\int \frac{\ln x}{x} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C.\]

Evaluate the Definite Integral

Use the result from Step 4 to evaluate the definite integral:\[\int_{1}^{b} \frac{\ln x}{x} \, dx = \left[ \frac{(\ln x)^2}{2} \right]_{1}^{b} = \frac{(\ln b)^2}{2} - \frac{(\ln 1)^2}{2}.\]Since \( \ln 1 = 0 \), this simplifies to \( \frac{(\ln b)^2}{2} \).

Determine the Limit

Find the limit as \( b \to \infty \):\[\lim_{b \to \infty} \frac{(\ln b)^2}{2} = \infty.\]This indicates that the integral diverges.

Key concepts

Convergence and Divergence

Improper integrals often need to be classified as either convergent or divergent. This classification helps us determine whether the integral has a finite value or not.
For the integral in question, we determine convergence or divergence by examining the behavior of the integral as the limit of integration approaches infinity.
If the value approaches a finite number, the integral converges. If it becomes infinite or fails to approach any value, it diverges.
In our example, the integral \( \int_{1}^{\infty} \frac{\ln x}{x} \, dx \) diverges because the limit goes to infinity, meaning it does not result in a finite value.

Limit of Integration

The limit of integration is crucial when dealing with improper integrals. It refers to the upper limit in an integral setup that involves infinity.
Specifically, for the integral \( \int_{1}^{\infty} \frac{\ln x}{x} \, dx \), the upper limit is \(\infty\), making this integral improper.
To handle this, we often transform the integral into a limit expression like \( \lim_{b \to \infty} \int_{1}^{b} \frac{\ln x}{x} \, dx \). This helps us examine what happens to the integral as the upper boundary tends towards infinity, and whether it yields a finite result.
This process is essential for determining convergence or divergence.

Indefinite Integral

An indefinite integral, unlike the definite one, doesn't have fixed limits. It results in a family of functions, commonly known as the antiderivative.
In this case, finding \( \int \frac{\ln x}{x} \, dx \) is necessary to evaluate the improper integral from 1 to \( \infty \).
Using substitution, we let \( u = \ln x \) which simplifies the problem, yielding an antiderivative of \( \frac{(\ln x)^2}{2} + C \). The constant \( C \) represents any constant since indefinite integrals provide a general solution.
This antiderivative is used to find the value of the improper integral by applying specific limits later on.

Substitution Method

The substitution method is an essential technique in solving integrals, especially when the integrand is complex or involves a compound function.
With the given integral \( \int \frac{\ln x}{x} \, dx \), substitution helps simplify the integrand to a more manageable form.
Here, substitution involves setting \( u = \ln x \), which leads to \( du = \frac{1}{x} \, dx \). This transforms the integral into \( \int u \, du \), a basic form that's easy to integrate, giving \( \frac{u^2}{2} + C \).
This method simplifies solving improper integrals significantly by reducing complicated expressions into simpler integrals that we know how to resolve.