Question

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.

If p > 1, then${\int }_1^{\infty }\frac{1}{x^p}dx$converges to$\frac{1}{p-1}.$

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information.

The given integral is${\int }_1^{\infty }\frac{1}{x^p}dx.$

Step 2. Prove. 

To prove if p > 1, then ${\int }_1^{\infty }\frac{1}{x^p}dx$converges to $\frac{1}{p-1},$ let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649067096801" $[a,\infty ),$then ${\int }_a^{\infty }f\left(x\right)dx=\underset{B\to \infty }{\lim }{\int }_a^{B}f\left(x\right)dx.$

So,

$$\begin{gathered} {\int }_1^{\infty }\frac{1}{x^p}dx \\ =\underset{B\to \infty }{\lim }{\int }_1^B{x}^{-p}dx \\ =\underset{B\to \infty }{\lim }{\left[\frac{x^{-p+1}}{-p+1}\right]}_1^B \\ =\underset{B\to \infty }{\lim }\left[\frac{B^{1-p}}{1-p}-\frac{1}{1-p}\right] \\ =0-\frac{1}{1-p} \\ =-\left(\frac{1}{-\left(p-1\right)}\right) \\ =\frac{1}{p-1} \end{gathered}$$

Now, p > 1, so the given integral converges to $\frac{1}{p-1}.$

Hence, the integral is proved.