Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.PE.65

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

Guida verificata passo dopo passo
1
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 \(t \to \infty\).

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
3m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

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.
Video consigliato:
Percorso guidato
11:11
Improper Integrals: Infinite Intervals

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.
Video consigliato:
Percorso guidato
05:44
Divergence Test (nth Term Test)

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.
Video consigliato:
5:26
Graphs of Logarithmic Functions