Chapter 5.10, Problem 2E

To determine

To find whether the integral ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{2x-1}dx}$ is improper or proper

Answer to Problem 2E

The given integral ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{2x-1}dx}$ is proper

Explanation of Solution

Given.

  ${\displaystyle \underset{1}{\overset{2}{\int }}\frac{1}{2x-1}dx}$

On integration, the integral becomes

  $\begin{array}{l}\text{ =}{\left[\frac{\ln (2x-1)}{2}\right]}_1^2\\ =\left[\frac{\ln (3)}{2}-0\right],\text{finite}\text{number}\end{array}$

Since the given integral has finite value.

Hence, the given integral is a proper integral

(b) 

To determine

To find whether the integral ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2x-1}dx}$ is improper or proper

Answer to Problem 2E

The given integral ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2x-1}dx}$ is improper

Explanation of Solution

Given,

The given function is discontinues at

  $x=\frac{1}{2}\text{, }$ the integral is splitted as:

  $\text{ =}{\displaystyle \underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1}{2x-1}dx+}{\displaystyle \underset{\frac{1}{2}}{\overset{1}{\int }}\frac{1}{2x-1}dx}$

Since the given integral has infinite value.

Hence, the given integral is improper

(c) 

To determine

To find whether the integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\sin x}{1+x^2}dx}$ is improper or proper

Answer to Problem 2E

The given integral ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\sin x}{1+x^2}dx}$ is improper

Explanation of Solution

Given,

  ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\sin x}{1+x^2}dx}$

In the interval $\text{[-}\infty ,\infty ],\frac{1}{1+x^2}>0$ , where as,

  $sinx<0\text{on [-}\infty ,0\text{] and }sinx>0\text{ on [0,}\infty \text{]}$ , the integral is splitted as:

  $\text{ =}{\displaystyle \underset{-\infty }{\overset{0}{\int }}\frac{\sin x}{1+x^2}dx+}{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{\sin x}{1+x^2}dx}$

Since, the solution of the above integral could not be evaluated by traditional methods of integration and have imaginary solution.

Hence, the given integral is improper

(d) 

To determine

To find whether the integral ${\displaystyle \underset{1}{\overset{2}{\int }}\ln (2x-1)dx}$ is improper or proper

Answer to Problem 2E

The given integral ${\displaystyle \underset{1}{\overset{2}{\int }}\ln (2x-1)dx}$ is proper

Explanation of Solution

Given,

  ${\displaystyle \underset{1}{\overset{2}{\int }}\ln (2x-1)dx}$

Integrating by parts, the above integral becomes as:

  $\begin{array}{l}={\left[\ln (2x-1).x\right]}_1^3-{\displaystyle \underset{1}{\overset{2}{\int }}\frac{2x}{2x-1}dx}\\ =\left[\ln (5).3-0\right]-{\displaystyle \underset{1}{\overset{2}{\int }}\frac{2x-1+1}{2x-1}dx}\\ =\ln (5).3-{\displaystyle \underset{1}{\overset{2}{\int }}\left(1+\frac{1}{2x-1}\right)dx}\\ $ =\ln (5).3-{[x+\frac{\ln (2x-1)}{2}]}_1^2$=\ln (5).3-[2+\frac{\ln (3)}{2}-1]\end{array}$

Since the integral has finite value.

Hence, the given integral is a proper integral