For the improper integral ∫_1^∞ 1/x^p dx with p > 1, what is the value when p = 2?

For the improper integral ∫_1^∞ 1/x^p dx with p > 1, what is the value when p = 2?

• A. 0
• B. ∞
• C. 2
• D. 1

Answer: (D)

For this improper integral, convergence happens when p>1 because the tail 1/x^p decreases fast enough to yield a finite area. In fact, for p>1 the value is 1/(p-1). With p = 2, that gives 1/(2-1) = 1. You can see this directly by evaluating the integral up to a variable upper limit and taking the limit: ∫_1^b x^{-2} dx = [-1/x]_1^b = -1/b + 1, and as b → ∞ this tends to 1.

For this improper integral, convergence happens when p>1 because the tail 1/x^p decreases fast enough to yield a finite area. In fact, for p>1 the value is 1/(p-1). With p = 2, that gives 1/(2-1) = 1. You can see this directly by evaluating the integral up to a variable upper limit and taking the limit: ∫_1^b x^{-2} dx = [-1/x]_1^b = -1/b + 1, and as b → ∞ this tends to 1.