Chapter 9.4, Problem 22E

To determine

To find: the value of improper integral.

Answer to Problem 22E

The value of integral is $0$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}$ .

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

  ${\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{0}2x{e}^{-x^2}dx+}\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^{a}2x{e}^{-x^2}dx}$

To find the value of integral, first integrate and then simplify,

  $\begin{array}{l}{\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{0}2x{e}^{-x^2}dx+}\underset{a\to \infty }{\lim }{\displaystyle {\int }_0^{a}2x{e}^{-x^2}dx}\\ {\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=\underset{a\to -\infty }{\lim }{\left[-e^{-x^2}\right]}_a^0\text{+}\underset{a\to \infty }{\lim }{\left[-e^{-x^2}\right]}_0^a\text{ [}{\displaystyle \int e^{x}dx=e^x\text{ and simplify}]}\\ {\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=\underset{a\to -\infty }{\lim }\left[-e^0+e^{-a^2}\right]\text{+}\underset{a\to \infty }{\lim }\left[-e^{-a^2}+e^0\right]\text{ [simplify]}\\ {\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=-1+\infty -\infty +1\text{ [simplify]}\\ {\displaystyle {\int }_{-\infty }^{\infty }2x{e}^{-x^2}dx}=0\end{array}$

Thus, the value is $0$ .