Question

The following improper integral $$ \int_1^\infty \frac{dx}{x^p} $$ converges for which value of $ p $?

• A. \( p>1 \)
• B. \( p<2 \)
• C. \( p \geq 0 \)
• D. \( p = 0 \)

Hint:

Improper integrals involving powers of \( x \) converge when the exponent is greater than 1. Always check the condition for convergence based on the exponent in the integrand.

Answer 1

The Correct Option is A

The improper integral is of the form: \[ I(p) = \int_1^\infty \frac{dx}{x^p} \] To determine the value of \( p \) for which this integral converges, we first evaluate the integral for different values of \( p \). The general formula for this type of integral is: \[ \int_1^\infty x^{-p} dx = \frac{1}{1-p} \quad \text{for} \quad p>1 \] For \( p \leq 1 \), the integral does not converge because the integral will diverge as \( x \to \infty \).
Therefore, the integral converges for \( p>1 \).
Thus, the correct answer is 1. \( p>1 \).