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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 8.8.81a

81. Find the values of p for which each integral converges.
a. ∫ from 1 to 2 of [dx / (x (ln x)^p)]

Guida verificata passo dopo passo
1
Identify the integral and the parameter involved: we have the integral \( \int_1^2 \frac{dx}{x (\ln x)^p} \), and we want to find for which values of \( p \) this integral converges.
Check the behavior of the integrand near the limits of integration. Since the interval is from 1 to 2, and \( \ln 1 = 0 \), the potential issue is at the lower limit \( x = 1 \) where the denominator involves \( (\ln x)^p \) which tends to zero, possibly causing a singularity.
Make a substitution to analyze the behavior near \( x = 1 \): let \( t = \ln x \). Then \( dt = \frac{1}{x} dx \), so \( dx = x dt = e^t dt \). The integral limits change from \( x=1 \) to \( t=0 \), and from \( x=2 \) to \( t=\ln 2 \).
Rewrite the integral in terms of \( t \): \( \int_0^{\ln 2} \frac{e^t dt}{e^t t^p} = \int_0^{\ln 2} \frac{dt}{t^p} \). Now the integral simplifies to \( \int_0^{\ln 2} t^{-p} dt \).
Determine convergence of \( \int_0^{a} t^{-p} dt \) for some positive \( a \). This integral converges if and only if \( -p > -1 \), or equivalently \( p < 1 \). So the original integral converges for \( p < 1 \).

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