Chapter 8.4, Problem 6E

To determine

To evaluate: The improper integral, ${\displaystyle {\int }_1^{\infty }\frac{2dx}{x^3}}$ or state that it diverges.

Answer to Problem 6E

The value of the integral is 1.

Explanation of Solution

Given information: The given integral is ${\displaystyle {\int }_1^{\infty }\frac{2dx}{x^3}}$ .

Concept used: Consider that $f\left(x\right)$ is continuous on $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)dx=\underset{b\to \infty }{\lim }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

Calculation:

Assume $I={\displaystyle {\int }_1^{\infty }\frac{2dx}{x^3}}$ . In order to express the improper integral as a limit definite integral, consider a real number $b$ such that,

  $I=\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{2dx}{x^3}}$

Apply the rules of integration to determine the value of integral in terms of $b$ .

  $\begin{array}{l}{\displaystyle {\int }_1^b\frac{2dx}{x^3}}={\left[-\frac{2}{2x^2}\right]}_1^b\\ \Rightarrow {\displaystyle {\int }_1^b\frac{2dx}{x^3}}={\left[-\frac{1}{x^2}\right]}_1^b\\ \Rightarrow {\displaystyle {\int }_1^b\frac{2dx}{x^3}}=\left(-\frac{1}{b^2}\right)-\left(-\frac{1}{{\left(1\right)}^2}\right)\\ \Rightarrow {\displaystyle {\int }_1^b\frac{2dx}{x^3}}=-\frac{1}{b^2}+1\end{array}$

Determine the value of $\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{2dx}{x^3}}$ .

  $\begin{array}{l}\underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{2dx}{x^3}}=\underset{b\to \infty }{\lim }\left[-\frac{1}{b^2}+1\right]\\ \Rightarrow \underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{2dx}{x^3}}=-\frac{1}{{\left(\infty \right)}^3}+1\\ \Rightarrow \underset{b\to \infty }{\lim }{\displaystyle {\int }_1^b\frac{2dx}{x^3}}=1\end{array}$

Since the given improper integral has some value, it converges to 1.