For what values of \( p \) does \( \int_{1}^{\infty} x^{-p} d x \) converge?

We start with the integral \[ \int_{1}^{\infty} x^{-p} \, dx. \] **Step 1. Evaluate the indefinite integral:** For \( p \neq 1 \), the antiderivative of \( x^{-p} \) is \[ \int x^{-p} \, dx = \frac{x^{-p+1}}{-p+1}. \] **Step 2. Write the definite integral using the antiderivative:** \[ \int_{1}^{b} x^{-p} \, dx = \left[ \frac{x^{-p+1}}{-p+1} \right]_{x=1}^{x=b} = \frac{b^{-p+1} - 1}{-p+1}. \] We are interested in the limit as \( b \to \infty \): \[ \lim_{b\to\infty} \frac{b^{-p+1} - 1}{-p+1}. \] **Step 3. Consider the behavior as \( b \to \infty \):** - If \( -p+1 < 0 \) (that is, if \( p > 1 \)), then \( b^{-p+1} \to 0 \) as \( b \to \infty \). The integral becomes \[ \frac{0 - 1}{-p+1} = \frac{1}{p-1}, \] which is finite. - If \( p = 1 \), the integrand becomes \( x^{-1} \), and the integral is \[ \int_{1}^{\infty} \frac{1}{x} \, dx, \] which is a divergent logarithmic integral. - If \( -p+1 > 0 \) (that is, if \( p < 1 \)), then \( b^{-p+1} \) grows without bound as \( b \to \infty \) and the integral diverges. **Conclusion:** The integral converges if and only if \( p > 1 \).