Chapter 5.10, Problem 51E

To determine

To find the value of $p$ for which the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{x^p}dx}$ converges.

Answer to Problem 51E

The integral is convergent, when $p$ is $\frac{1}{1-p}$

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{x^p}dx}$

Concept used: The integral is convergent, if the integral has finite value.

Formula used:

 i ${\displaystyle \int x^{n}dx=}\frac{x^{n+1}}{n+1}$

 ii ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{x^p}dx\text{is}\text{convergent,}\text{when}p>1\text{else}\text{it}\text{is}\text{divergent}}$

Calculation :

The integral becomes as: ${\left[\frac{x^{-p+1}}{-p+1}\right]}_0^1=\frac{1}{1-p}{[x^{-p+1}]}_0^1$

Case I: When $p>1\text{or}p-1>0$

The integral becomes as: $\frac{1}{1-p}{\left[\frac{1}{x^{(p-1)}}\right]}_0^1$ , which is an infinite number.

For $p>1,$ , integral is divergent

Case II: When $p<1\text{or}p-1<0$

The integral becomes as: $\frac{1}{1-p}{\left[x^{-(p-1)}\right]}_0^1=\frac{1}{1-p}$ , which is a finite number.

For $p<1,$ , integral is convergent and the value of $p$ is $\frac{1}{1-p}$

Case III: When $p=1$

The integral becomes as: $\frac{1}{1-p}{\left[x^{-(p-1)}\right]}_0^1$ , which is an infinite number.

Therefore, the value of $p$ is $\frac{1}{1-p}$ for which ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{x^p}dx}$ is convergent.