Chapter 9.4, Problem 3E

a.

To determine

To find: the expression that represents the improper integral as a limit of a definite integrals.

Answer to Problem 3E

The expression that represents the improper integral as a limit of definite integrals is $\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0\frac{2x}{{\left(x^2+1\right)}^2}dx+}\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0\frac{2x}{{\left(x^2+1\right)}^2}dx+}\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{2x}{{\left(x^2+1\right)}^2}dx}$

b.

To determine

To find: the value of integral.

Answer to Problem 3E

The value of integral is $0$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0\frac{2x}{{\left(x^2+1\right)}^2}dx+}\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{2x}{{\left(x^2+1\right)}^2}dx}$

Now, evaluate the definite integrals first integrate then use limit,

Thus, the integral is $0$ .