Chapter 9.4, Problem 58E

To determine

To state: a convincing argument to satisfy the theorem $6$ .

Answer to Problem 58E

On the interval $\text{[}\pi \text{,}\infty \text{)}$ and the integral ${\displaystyle {\int }_{\pi }^{\infty }\frac{1}{x}}$ diverges,

  $0\le \frac{1}{x}\le \frac{2+\cos x}{x}$

Thus, the ${\displaystyle {\int }_{\pi }^{\infty }\frac{2+\cos x}{x}dx}$ diverges.

On the interval $\text{[1,}\infty \text{)}$ and the integral ${\displaystyle {\int }_1^{\infty }\frac{1}{x^3}}$ converges,

  $0\le \frac{1}{1+x^3}\le \frac{1}{x^3}$

Thus, the ${\displaystyle {\int }_1^{\infty }\frac{1}{x^3+1}dx}$ converges.

Explanation of Solution

Given information:

Theorem $6$ : let $f\text{ and }g$ be continuous on $[a,\infty )$ with $0\le f\left(x\right)\le g\left(x\right)$ for all $x\ge a$ ,

1. ${\displaystyle {\int }_0^{\infty }f\left(x\right)dx}$ converges if ${\displaystyle {\int }_0^{\infty }g\left(x\right)dx}$converges.
2. ${\displaystyle {\int }_0^{\infty }g\left(x\right)dx}$ diverges if ${\displaystyle {\int }_0^{\infty }f\left(x\right)dx}$ diverges.

On the interval $\text{[}\pi \text{,}\infty \text{)}$ and the integral ${\displaystyle {\int }_{\pi }^{\infty }\frac{1}{x}}$ diverges,

  $0\le \frac{1}{x}\le \frac{2+\cos x}{x}$

Thus, the ${\displaystyle {\int }_{\pi }^{\infty }\frac{2+\cos x}{x}dx}$ diverges.

On the interval $\text{[1,}\infty \text{)}$ and the integral ${\displaystyle {\int }_1^{\infty }\frac{1}{x^3}}$ converges,

  $0\le \frac{1}{1+x^3}\le \frac{1}{x^3}$

Thus, the ${\displaystyle {\int }_1^{\infty }\frac{1}{x^3+1}dx}$ converges.