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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 8 - Integration TechniquesProblema 8.R.92
Capitolo 8, Problema 8.R.92

92. Integral with a parameter For what values of p does the integral
∫ (from 1 to ∞) dx/xlnᵖ(x) converge, and what is its value (in terms of p)?

Guida verificata passo dopo passo
1
First, rewrite the integral to clearly identify the integrand and the limits: consider the integral \( \int_1^{\infty} \frac{1}{x (\ln x)^p} \, dx \).
To analyze convergence, perform the substitution \( t = \ln x \). Then, \( dt = \frac{1}{x} dx \), which implies \( dx = x \, dt \). Substitute into the integral:
\[ \int_1^{\infty} \frac{1}{x (\ln x)^p} \, dx = \int_0^{\infty} \frac{1}{t^p} \, dt \]
Now, examine the integral \( \int_0^{\infty} t^{-p} \, dt \) for convergence. Since the integral is improper at both limits, analyze the behavior near 0 and near infinity separately:
- Near \( t = 0 \), the integral \( \int_0^1 t^{-p} \, dt \) converges if and only if \( p < 1 \).
- Near \( t = \infty \), the integral \( \int_1^{\infty} t^{-p} \, dt \) converges if and only if \( p > 1 \).
Since both conditions cannot be true simultaneously, the integral converges only if \( p > 1 \) because the original integral's lower limit corresponds to \( t = 0 \) and the upper limit to \( t = \infty \). Actually, the substitution shows the lower limit is \( t=0 \), so we must be careful: the integral converges if \( p > 1 \) because the integral near infinity converges only then, and near zero the integral converges for \( p < 1 \), but since the lower limit is 0, the integral diverges there if \( p \geq 1 \).
Finally, for \( p > 1 \), compute the value of the integral by evaluating \( \int_0^{\infty} t^{-p} \, dt \) with proper limits and express the result in terms of \( p \).

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Improper Integrals and Convergence

Improper integrals involve integration over infinite intervals or integrands with unbounded behavior. To determine convergence, one must analyze the limit of the integral as the upper bound approaches infinity. If this limit exists and is finite, the integral converges; otherwise, it diverges.
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Improper Integrals: Infinite Intervals

Behavior of the Integrand Near Infinity

Understanding how the function behaves as x approaches infinity is crucial. For the integral ∫₁^∞ dx / (x ln^p(x)), the decay rate of the denominator, especially the power p on the logarithm, determines whether the integral converges. Slower decay (small p) may lead to divergence, while faster decay (larger p) can ensure convergence.
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Completing the Square to Rewrite the Integrand

Use of Substitution in Evaluating Integrals

Substitution simplifies integrals by changing variables to a more manageable form. For this integral, substituting t = ln(x) transforms the integral into one involving t, making it easier to analyze convergence and compute the integral's value in terms of p.
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Substitution With an Extra Variable
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