Determine for which values of the parameter p the improper integral ∫_1^(∞)dx/x^p converges, and give its value when it converges.

Convergent for p>1; value =1/(p-1)

Determine for which values of the parameter $p$ the improper integral $\int_{1}^{\infty} \frac{d x}{x ^{p}}$ converges, and give its value when it converges.

hard 4 marks Class 12 · Mathematics · Integrals

Answer

Convergent for

p > 1

; value

= \frac{1}{p - 1}

The correct answer is Convergent for $p > 1$ ; value $= \frac{1}{p - 1}$ — Thus the integral converges exactly for $p > 1$ and its value is $\frac{1}{p - 1}$ .

Step-by-step solution

Write the improper integral as a limit: $\int_{1}^{\infty} \frac{d x}{x ^{p}} = lim_{b \to \infty} \int_{1}^{b} x^{- p} d x$ .
Integrate $x^{- p}$ for $p \neq = 1$ : $\int x^{- p} d x = \frac{x ^{1 - p}}{1 - p}$ .
Apply the limits: $lim_{b \to \infty} [\frac{b ^{1 - p}}{1 - p} - \frac{1 ^{1 - p}}{1 - p}]$ .
Analyze $b^{1 - p}$ as $b \to \infty$ : it tends to $0$ only if $1 - p < 0$ , i.e., $p > 1$ .
If $p > 1$ , the limit becomes $- \frac{1}{1 - p} = \frac{1}{p - 1}$ .
If $p \leq 1$ , $b^{1 - p}$ diverges, so the integral does not converge.
Thus the integral converges exactly for $p > 1$ and its value is $\frac{1}{p - 1}$ .

Common mistake: Assuming the integral converges for

p \leq 1

, especially forgetting the case

p = 1

where the integral becomes

\int_{1}^{\infty} \frac{d x}{x}

which diverges logarithmically.

Concept tested: convergence of improper integrals

Published 31 August 2026 · How this solution was prepared and checked

Determine for which values of the parameter p the improper integral ∫1∞​xpdx​ converges, and give its value when it converges.

Step-by-step solution

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Determine for which values of the parameter $p$ the improper integral $\int_{1}^{\infty} \frac{d x}{x ^{p}}$ converges, and give its value when it converges.