Problem 4 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The integral converges for \(p > 1\).

Step 1: Find an antiderivative

We need to find an antiderivative for \(x^{-p}\) in order to evaluate the integral. The antiderivative of \(x^{-p}\) is \(\frac{x^{-p+1}}{-p+1}\), which we can verify by taking the derivative, assuming \(p \neq 1\).

Step 2: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we can now evaluate the integral: $$\int_{1}^{\infty} x^{-p} d x = \lim_{b \to \infty} \left[ \frac{x^{-p+1}}{-p+1} \right]_1^b,$$ where we apply the limit since it is an improper integral.

Step 3: Evaluate the limit

Taking the limit as b approaches infinity, we get: $$\lim_{b \to \infty} \left[ \frac{b^{-p+1}}{-p+1} - \frac{1^{-p+1}}{-p+1} \right].$$

Step 4: Determine when the limit converges

To make the expression converge, we want the first term of the expression to approach 0: $$\lim_{b \to \infty} \frac{b^{-p+1}}{-p+1}.$$ Notice that this limit converges to 0 if \(-p+1 < 0\) or \(p > 1\). Therefore, if \(p > 1\), the integral converges and diverges otherwise. So, the integral \(\int_{1}^{\infty} x^{-p} d x\) converges for \(p > 1\).