Chapter 8.4, Problem 25E

(a)

To determine

To state: Why the given integral is improper.

Answer to Problem 25E

It has an infinite discontinuity at the point $x=1$ .

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{dx}{1-x^2}}$.

It is known that the integral is said to be an improper integral if the integral function becomes at any point which lies in the interval of integration.

Since the given integral has an infinite discontinuity at the point $x=1$ , the given integral is an improper integral.

(b)

To determine

To evaluate: The given improper integral or state that it diverges.

Answer to Problem 25E

The improper integral is $\infty$ , the given improper integral diverges.

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{dx}{1-x^2}}$

Concept Used: Consider that function $f\left(x\right)$ is continuous on $\left[a,c\right)\cup \left(c,b\right]$ , then ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx=}{\displaystyle \underset{a}{\overset{c}{\int }}f\left(x\right)dx}+{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)dx}$ .

Calculation:

The given integral has an infinite discontinuity at the point $x=1$ .

So,

  ${\displaystyle {\int }_0^2\frac{dx}{1-x^2}}={\displaystyle {\int }_0^1\frac{dx}{1-x^2}}+{\displaystyle {\int }_1^2\frac{dx}{1-x^2}}$

Now, determine each of the obtained improper integral which is obtained on the right side of the obtained equation.

  $\begin{array}{c}{\displaystyle {\int }_0^2\frac{dx}{1-x^2}}=\underset{c\to 1^{-}}{\lim }{\displaystyle {\int }_0^c\frac{dx}{1-x^2}}+\underset{c\to 1^{+}}{\lim }{\displaystyle {\int }_c^2\frac{dx}{1-x^2}}\\ =\underset{c\to 1^{-}}{\lim }{\displaystyle {\int }_0^c\frac{dx}{\left(1-x\right)\left(1+x\right)}}+\underset{c\to 1^{+}}{\lim }{\displaystyle {\int }_c^2\frac{dx}{\left(1-x\right)\left(1+x\right)}}\end{array}$

Use the partial fraction to integrate the obtained integral which is shown below.

  $\begin{array}{c}\frac{1}{\left(1-x\right)\left(1+x\right)}=\frac{A}{1-x}+\frac{B}{1+x}\\ \frac{1}{\left(1-x\right)\left(1+x\right)}=\frac{A\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}+\frac{B\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\\ =\frac{\left(A-B\right)x+A+B}{\left(1-x\right)\left(1+x\right)}\end{array}$

Compare the coefficient of $x$ and constant, and solve for $A$ and $B$ .

  $A=B=\frac{1}{2}$

So,

  $\begin{array}{c}\frac{1}{\left(1-x\right)\left(1+x\right)}=\frac{1}{2\left(1-x\right)}+\frac{1}{2\left(1+x\right)}\\ =\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\end{array}$

So,

  $\begin{array}{c}{\displaystyle {\int }_0^2\frac{dx}{1-x^2}}=\underset{c\to 1^{-}}{\lim }{\displaystyle {\int }_0^c\frac{dx}{\left(1-x\right)\left(1+x\right)}}+\underset{c\to 1^{+}}{\lim }{\displaystyle {\int }_c^2\frac{dx}{\left(1-x\right)\left(1+x\right)}}\\ =\underset{c\to 1^{-}}{\lim }\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)+\underset{c\to 1^{+}}{\lim }\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right)\\ =\infty \end{array}$

Since the obtained result is $\infty$ , the given improper integral diverges.