Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

${\int }_0^{\infty }x{e}^{-x^2}dx$

Short answer

The improper integral on $\left(0,\infty \right)$converges.

${\int }_0^{\infty }x{e}^{-x^2}dx=\frac{1}{2}$

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

${\int }_0^{\infty }x{e}^{-x^2}dx$

Step 2. Value of integral. 

Solve the integral by using the definition of improper integral.

$$\begin{gathered} {\int }_0^{\infty }x{e}^{-x^2}dx=\underset{a\to \infty }{\lim }{\int }_0^{a}x{e}^{-x^2}dx \\ =\underset{a\to \infty }{\lim }{\left[-\frac{1}{2}{e}^{-x^2}\right]}_0^a \\ =\underset{a\to \infty }{\lim }\left[-\frac{1}{2}{e}^{-a^2}-\left(\frac{-1}{2}\right)\right] \\ =0+\frac{1}{2} \\ =\frac{1}{2} \end{gathered}$$

The value of the integral is${\int }_0^{\infty }x{e}^{-x^2}dx=\frac{1}{2}$