Which of the improper integrals in Exercises 63–68 converge and which diverge?
∫ from 1 to ∞ of ((ln z) / z) dz

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Identify the integral to analyze: \(\displaystyle \int_1^{\infty} \frac{\ln z}{z} \, dz\).

Recognize that this is an improper integral because the upper limit of integration is infinite.

Consider the behavior of the integrand \(\frac{\ln z}{z}\) as \(z \to \infty\). To determine convergence, analyze the limit of the integral or use a comparison test.

Use substitution to simplify the integral: let \(t = \ln z\), which implies \(z = e^t\) and \(dz = e^t dt\). Rewrite the integral in terms of \(t\).

After substitution, express the integral with new limits and integrand, then analyze whether the resulting integral converges or diverges by evaluating the limit as \(t \to \infty\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, limits are used to define the integral as a limit of definite integrals over finite intervals. Determining convergence or divergence depends on whether this limit exists and is finite.

Convergence and Divergence of Integrals

An improper integral converges if the limit defining it exists and is finite; otherwise, it diverges. Testing convergence often involves comparison tests or evaluating the behavior of the integrand as the variable approaches infinity or a point of discontinuity.

Behavior of Logarithmic Functions in Integrals

Logarithmic functions like ln(z) grow slowly as z approaches infinity. When combined with other functions, such as 1/z, their growth rate affects the convergence of the integral. Understanding how ln(z)/z behaves for large z is key to determining if the integral converges.

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