Chapter 8.4, Problem 59E

(a)

To determine

To show: ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ converges for $n=0,1,2$ .

Explanation of Solution

Given information: The integral $f\left(n+1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ is given where $n\ge 0$ .

Concept used :

If $f\left(x\right)$ is continuous on the interval $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ .

Proof:

Substitute $n=0,1,2$ separately into the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ and simplify.

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^0}{e}^{-x}dx={\displaystyle \underset{0}{\overset{\infty }{\int }}1}{e}^{-x}dx\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-x}}dx\\ =\underset{c\to \infty }{\lim }{\displaystyle \underset{0}{\overset{c}{\int }}{e}^{-x}}dx\\ =\underset{c\to \infty }{\lim }{\left[-e^{-x}\right]}_0^c\\ =\underset{c\to \infty }{\lim }\left[-\left(e^{-c}-e^{-0}\right)\right]\\ =\underset{c\to \infty }{\lim }\left[-\left(e^{-c}-1\right)\right]\\ =\left[-\left(e^{-\infty }-1\right)\right]\\ =\left[-\left(0-1\right)\right]\\ =1\end{array}$

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^1}{e}^{-x}dx={\displaystyle \underset{0}{\overset{\infty }{\int }}x}{e}^{-x}dx\\ =\underset{c\to \infty }{\lim }{\displaystyle \underset{0}{\overset{c}{\int }}x}{e}^{-x}dx\end{array}$

Consider $u=x$ such that $du=dx$ and $dv=e^{-x}dx$ such that $v={\displaystyle \int e^{-x}}dx=-e^{-x}$ . $dv=e^{-x}dx$ Simplify the integral ${\displaystyle \underset{0}{\overset{c}{\int }}x}{e}^{-x}dx$ using integration by parts, that is, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ by substituting $u=x$ , $dx=e^{x}dv$ , and $v=-e^{-x}$ into ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{c}{\int }}x{e}^{-x}}d\theta ={\left[-x{e}^{-x}\right]}_0^c-{\displaystyle \underset{0}{\overset{c}{\int }}{e}^{-x}}dx\\ =\left[-c{e}^{-c}+0\right]-{\left[e^{-x}\right]}_0^c\\ =-c{e}^{-c}-e^{-c}+e^{-0}\\ =-c{e}^{-c}-e^{-c}+1\end{array}$

Therefore,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}x}{e}^{-x}dx=\underset{c\to \infty }{\lim }{\displaystyle \underset{0}{\overset{c}{\int }}x}{e}^{-x}dx\\ =\underset{c\to \infty }{\lim }\left[-c{e}^{-c}-e^{-c}+1\right]\\ =\left[-\left(\infty \right)e^{-\infty }-e^{-\infty }+1\right]\\ =\left[-0-0+1\right]\\ =1\end{array}$

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2}{e}^{-x}dx={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2}{e}^{-x}dx\\ =\underset{c\to \infty }{\lim }{\displaystyle \underset{0}{\overset{c}{\int }}{x}^2}{e}^{-x}dx\end{array}$

Consider $u=x^2$ such that $du=2xdx$ and $dv=e^{-x}dx$ such that $v={\displaystyle \int e^{-x}}dx=-e^{-x}$ . Simplify the integral ${\displaystyle \underset{0}{\overset{c}{\int }}{x}^2}{e}^{-x}dx$ using integration by parts, that is, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ by substituting $u=x^2$ , $du=2xdx$ , and $v=-e^{-x}$ into ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{c}{\int }}{x}^2{e}^{-x}}dx={\left[-x^2{e}^{-x}\right]}_0^c-{\displaystyle \underset{0}{\overset{c}{\int }}-2x{e}^{-x}}dx\\ ={\left[-x^2{e}^{-x}\right]}_0^c+{\displaystyle \underset{0}{\overset{c}{\int }}2x{e}^{-x}}dx\\ ={\left[-x^2{e}^{-x}\right]}_0^c+{\left[-x{e}^{-x}-e^{-x}\right]}_0^c\\ =\left[-c^2{e}^{-c}+0\right]+2\left[-c{e}^{-c}-e^{-c}+0e^{-0}+e^{-0}\right]\\ =-c^2{e}^{-c}-2c{e}^{-c}-2e^{-c}+2\end{array}$

Therefore,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2}{e}^{-x}dx=\underset{c\to \infty }{\lim }{\displaystyle \underset{0}{\overset{c}{\int }}{x}^2}{e}^{-x}dx\\ =\underset{c\to \infty }{\lim }\left[-c^2{e}^{-c}-2c{e}^{-c}-2e^{-c}+2\right]\\ =\underset{c\to \infty }{\lim }\left[e^{-c}\left(-c^2-2c-2\right)+2\right]\\ =0+2\\ =2\end{array}$

Since it is evident from the obtained calculations that ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ for $n=0,1,2$ is finite, the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ converges for $n=0,1,2$ .

Hence, it has been shown that the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ converges for $n=0,1,2$ .

(b)

To determine

To show: $f\left(n+1\right)=f\left(n\right)$ by suing integration by parts, where $f\left(n+1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ .

Explanation of Solution

Given information: The integral $f\left(n+1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ is given where $n\ge 0$

Concept used :

If $f\left(x\right)$ is continuous on the interval $\left[a,\infty \right)$ , then ${\displaystyle \underset{a}{\overset{\infty }{\int }}f\left(x\right)}dx=\underset{b\to \infty }{\lim }{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ .

Proof:

Consider $u=x^n$ such that $du=n{x}^{n-1}dx$ and $dv=e^{-x}dx$ such that $v={\displaystyle \int e^{-x}}dx=-e^{-x}$ . Simplify the integral ${\displaystyle \underset{0}{\overset{c}{\int }}{x}^n}{e}^{-x}dx$ using integration by parts, that is, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ by substituting $u=x^n$ , $du=n{x}^{n-1}dx$ , $dv=e^{-x}dx$ , and $v=-e^{-x}$ into ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ .

  $\begin{array}{c}{\displaystyle \int x^n{e}^{-x}dx}=-x^n{e}^{-x}-{\displaystyle \int -n{x}^{n-1}{e}^{-x}}dx\\ =-x^n{e}^{-x}+{\displaystyle \underset{0}{\overset{c}{\int }}n{x}^{n-1}{e}^{-x}}dx\end{array}$

Therefore,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-x}dx}={\left[-x^n{e}^{-x}\right]}_0^{\infty }+{\displaystyle \underset{0}{\overset{\infty }{\int }}n{x}^{n-1}{e}^{-x}}dx\\ =\underset{c\to \infty }{\lim }{\left[-x^n{e}^{-x}\right]}_0^c+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =\underset{c\to \infty }{\lim }\left[-c^n{e}^{-c}\right]+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =-\underset{c\to \infty }{\lim }\left[c^n\frac{1}{e^c}\right]+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =-\underset{c\to \infty }{\lim }\left(\frac{c^n}{e^c}\right)+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =-\underset{c\to \infty }{\lim }\left(e^{n\ln \left(c\right)-c\ln \left(e\right)}\right)+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =0+n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\\ =n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx\end{array}$

Since $f\left(n+1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ , $f\left(n\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}}{e}^{-x}dx$ .

Substitute ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}}{e}^{-x}dx=f\left(n\right)$ and ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx=f\left(n+1\right)$ into ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n{e}^{-x}dx}=n{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{n-1}{e}^{-x}}dx$ to get the required result.

  $f\left(n+1\right)=nf\left(n\right)$

Therefore, it has been proved that $f\left(n+1\right)=nf\left(n\right)$ .

(c)

To determine

To give: An argument that the integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ converges for all integers, where $n\ge 0$

Explanation of Solution

Given information: The integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ is given where $n\ge 0$

Formula used:

  $f\left(n+1\right)=nf\left(n\right)$

Calculation:

Since $f\left(n+1\right)=nf\left(n\right)$ , by induction $f\left(n+1\right)=n\left(n-1\right)\dots 1f\left(1\right)=n!$

  $n!$ is finite and hence $f\left(n+1\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^n}{e}^{-x}dx$ converges for all integers, where $n\ge 0$