Chapter 9.4, Problem 59E

a.

To determine

Toshow: that ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}$ converges for $n=0,1,2$ .

Explanation of Solution

Given information:

The function is $f\left(n+1\right)={\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dt}$ .

Proof :to show the integral is converges, first substitute the value of n and then integrate and simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}={\displaystyle {\int }_0^{\infty }{x}^0{e}^{-x}dx}\text{ [substitute }n=0\text{]}\\ {\displaystyle {\int }_0^{\infty }{x}^0{e}^{-x}dx}={\left[-e^{-x}\right]}_0^{\infty }\text{ [}{\displaystyle \int e^x}=e^x]\\ {\displaystyle {\int }_0^{\infty }{x}^0{e}^{-x}dx}=-e^{-\infty }+e^{-0}\\ {\displaystyle {\int }_0^{\infty }{x}^0{e}^{-x}dx}=1\text{ [simplify]}\end{array}$

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}={\displaystyle {\int }_0^{\infty }{x}^1{e}^{-x}dx}\text{ [substitute }x=1\text{]}\\ {\displaystyle {\int }_0^{\infty }{x}^1{e}^{-x}dx}={\left[-e^{-x}\left(x+1\right)\right]}_0^{\infty }\text{ [}{\displaystyle \int uvdx}=u{\displaystyle \int vdx}-{\displaystyle \int \frac{du}{dx}{\displaystyle \int vdxdx]}}\\ {\displaystyle {\int }_0^{\infty }{x}^1{e}^{-x}dx}=-e^{-\infty }\left(\infty +1\right)+e^{-0}\left(0+1\right)\\ {\displaystyle {\int }_0^{\infty }{x}^1{e}^{-x}dx}=1\text{ [simplify]}\end{array}$

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}={\displaystyle {\int }_0^{\infty }{x}^2{e}^{-x}dx}\text{ [substitute }n=2\text{]}\\ {\displaystyle {\int }_0^{\infty }{x}^2{e}^{-x}dx}={\left[-x^2{e}^{-x}-2e^{-x}\left(x+1\right)\right]}_0^{\infty }\text{ [}{\displaystyle \int uvdx}=u{\displaystyle \int vdx}-{\displaystyle \int \frac{du}{dx}{\displaystyle \int vdxdx]}}\\ {\displaystyle {\int }_0^{\infty }{x}^2{e}^{-x}dx}=-{\left(\infty \right)}^2{e}^{-\infty }-e^{-\infty }\left(\infty +1\right)+{\left(0\right)}^2{e}^{-0}+e^{-0}\left(0+1\right)\\ {\displaystyle {\int }_0^{\infty }{x}^2{e}^{-x}dx}=2\text{ [simplify]}\end{array}$

Thus, the integral is converges.

b.

To determine

To show: that $f\left(n+1\right)=nf\left(n\right)$ .

Explanation of Solution

Given information:

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

Proof: integration by parts the function can be defined as:

  ${\displaystyle \int x^n{e}^{-x}}dx=-x^n{e}^{-x}+n{\displaystyle \int x^{n-1}{e}^{-x}dx}+c$

Since, the term $-x^n{e}^{-x}$ has value zero at the point $x=0$ and has limit equal to zero as $x\to \infty$ , when the above equation is evaluated “from zero to infinity”.

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

c.

To determine

To state: a convincing argument that ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}$ converges for all integers $n\ge 0$ .

Answer to Problem 59E

the integral of ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}$ is follows the formula $f\left(n+1\right)=nf\left(n\right)$ by an induction argument.

In fact, it follows that ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}=n!$ .

Explanation of Solution

Given information:

  $f\left(n+1\right)={\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dt}$

the integral of ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}$ is follows the formula $f\left(n+1\right)=nf\left(n\right)$ by an induction argument.

In fact, it follows that ${\displaystyle {\int }_0^{\infty }{x}^n{e}^{-x}dx}=n!$ .