Chapter 9.4, Problem 54E

a.

To determine

Tofind: the value of the integral when $p=0.5$ .

Answer to Problem 54E

the integral is divergewhen $p=0.5$ .

Explanation of Solution

Given information:

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

Calculation: to find the value of the integral, first substitute the value then integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}={\displaystyle {\int }_1^{\infty }\frac{dx}{x^{0.5}}\text{ [substitute]}}\\ {{\displaystyle {\int }_1^{\infty }\frac{dx}{x^{0.5}}=\left[2\sqrt{x}\right]}}_1^{\infty }\text{ [}{\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}]\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^{0.5}}=2\sqrt{\infty }-2\sqrt{1}}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^{0.5}}=\text{not defined [simplify]}}\end{array}$

Thus, the integral is diverge.

b.

To determine

To find: the value of the integral when $p=1$ .

Answer to Problem 54E

the integral is divergewhen $p=1$ .

Explanation of Solution

Given information:

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

Calculation: to find the value of the integral, first substitute the value then integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}={\displaystyle {\int }_1^{\infty }\frac{dx}{x^1}\text{ [substitute]}}\\ {{\displaystyle {\int }_1^{\infty }\frac{dx}{x}=\left[\ln x\right]}}_1^{\infty }\text{ [}{\displaystyle \int \frac{1}{x}dx}=\ln x]\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x}=\ln \infty -\ln 1}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x}=\text{not defined [simplify]}}\end{array}$

Thus, the integral is diverge.

c.

To determine

To find: the value of the integral when $p=1.5$ .

Answer to Problem 54E

the value of integral is $-2$ when $p=1.5$ .

Explanation of Solution

Given information:

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

Calculation: to find the value of the integral, first substitute the value then integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}={\displaystyle {\int }_1^{\infty }\frac{dx}{x^{1.5}}\text{ [substitute]}}\\ {{\displaystyle {\int }_1^{\infty }\frac{dx}{x^{1.5}}=\left[\frac{2}{\sqrt{x}}\right]}}_1^{\infty }\text{ [}{\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}]\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^{1.5}}=\frac{2}{\sqrt{\infty }}-\frac{2}{\sqrt{1}}}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^{1.5}}=0-2\text{ [simplify]}}\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^{1.5}}}=-2\end{array}$

Thus, the value of integral is $-2$ .

d.

To determine

To show: that ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}=\underset{h\to \infty }{\lim }\left[\frac{1}{1-p}\left(\frac{1}{b^{p-1}}-1\right)\right]$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}=\underset{b\to \infty }{\lim }\left[\frac{1}{1-p}\left(\frac{1}{b^{p-1}}-1\right)\right]$ .

Proof :to show that integrate first then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }}{\displaystyle {\int }_1^h\frac{dx}{x^{1.5}}}\\ {{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }\left[\frac{x^{-p+1}}{-p+1}\right]}}_1^b\text{ [}{\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}]\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }}\left(\frac{1}{1-p}\left(\frac{1}{b^{-p+1}}-\frac{1}{1^{-p+1}}\right)\right)\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }\left(\frac{1}{1-p}\left(\frac{1}{b^{-p+1}}-1\right)\right)\text{ [simplify]}}\end{array}$

e.

To determine

To show: that ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}=\left\{\begin{array}{l}\frac{1}{p-1},p>1\\ \infty ,p<1\end{array}\right\}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}=\left\{\begin{array}{l}\frac{1}{p-1},p>1\\ \infty ,p<1\end{array}\right\}$ .

Proof :to show that integrate first then simplify,

  $\begin{array}{l}{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }}{\displaystyle {\int }_1^h\frac{dx}{x^{1.5}}}\\ {{\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }\left[\frac{x^{-p+1}}{-p+1}\right]}}_1^b\text{ [}{\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}]\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }}\left(\frac{1}{1-p}\left(\frac{1}{b^{-p+1}}-\frac{1}{1^{-p+1}}\right)\right)\\ {\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}=\underset{b\to \infty }{\lim }\left(\frac{1}{1-p}\left(\frac{1}{b^{-p+1}}-1\right)\right)\text{ [simplify]}}\end{array}$

Now for $p>1$ ,

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

For $p<1$ ,

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

e.

To determine

To find: the values for the integral is converge and diverge.

Answer to Problem 54E

for $p>1$ the integral is converges and for $p<1$ the integral is diverge.

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_1^{\infty }\frac{dx}{x^p}}=\left\{\begin{array}{l}\frac{1}{p-1},p>1\\ \infty ,p<1\end{array}\right\}$ .

Calculation : from the above part it can be observed that, for $p>1$ the integral is converges and for $p<1$ the integral is diverge.