Question

Convergence and divergence of basic improper integrals: Determine whether each of the given improper integrals converges or diverges. For those that converge, give the exact solution of the integral.

${\int }_1^{\infty }\frac{1}{x^p}dx,p=1$

Short answer

The integral is divergent.

Step-by-step solution

Step 1. Given information.

Consider the given information.

${\int }_1^{\infty }\frac{1}{x^p}dx$

Step 2. Solve the integral.

Solve the given integral by applying the power rule.

$\begin{array}{rcl}{\int }_1^{\infty }\frac{1}{x^p}dx& =& {\int }_1^{\infty }{x}^{-p}dx\\ & =& {\left[\frac{x^{-p+1}}{-p+1}\right]}_1^{\infty }\\ & =& \frac{1}{1-p}\left[{\infty }^{-p+1}-1^{-p+1}\right]\\ & =& \mathrm{Undefined}\end{array}$

The intergal is divegent for p=1.