Question

How do you determine if the improper integral converges or diverges $\int\limits_{0}^{\infty }{{{x}^{2}}{{e}^{-x}}dx}$?

Answer 1

We first convert the integral form to limit form where $\int\limits_{0}^{\infty }{{{x}^{2}}{{e}^{-x}}dx}=\displaystyle \lim_{k \to \infty }\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}$. We find the limit value of the integral using parts form. If there is a limit value then we have a converging integral.

Complete step-by-step answer:
We will change the integral form to limit form.
We will take the upper limit as $k$ whose condition is $k \to \infty$.
Therefore, $\int\limits_{0}^{\infty }{{{x}^{2}}{{e}^{-x}}dx}=\displaystyle \lim_{k \to \infty }\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}$.
We now try to find the definite integral form of $\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}$.
We need to find the integration of $\int{{{x}^{2}}{{e}^{-x}}dx}$ using integration by parts method.
Integration by parts method is usually used for the multiplication of the functions and their integration.
Let’s assume $f\left( x \right)=g\left( x \right)h\left( x \right)$. We need to find the integration of $\int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx}$.
We take $u=g\left( x \right),v=h\left( x \right)$. This gives $\int{f\left( x \right)dx}=\int{uvdx}$.
The theorem of integration by parts gives $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}$.
For our integration $\int{{{x}^{2}}{{e}^{-x}}dx}$, we take $u={{x}^{2}},v={{e}^{-x}}$.
Now we complete the integration $\int{{{x}^{2}}{{e}^{-x}}dx}={{x}^{2}}\int{{{e}^{-x}}dx}-\int{\left( \dfrac{d\left( {{x}^{2}} \right)}{dx}\int{{{e}^{-x}}dx} \right)dx}$.
We have the differentiation formula for $u={{x}^{2}}$ where $\dfrac{du}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$.
The integration formula for $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$.
We apply these formulas to complete the integration and get
$\int{{{x}^{2}}{{e}^{-x}}dx}={{x}^{2}}\left( -{{e}^{-x}} \right)+\int{\left( 2x\times {{e}^{-x}} \right)dx}=-{{x}^{2}}{{e}^{-x}}+2\int{x{{e}^{-x}}dx}$.
We again apply by parts for $\int{x{{e}^{-x}}dx}$
So, $\int{x{{e}^{-x}}dx}=x\left( -{{e}^{-x}} \right)+\int{{{e}^{-x}}dx}=-x{{e}^{-x}}-{{e}^{-x}}$.
The final integral form is

& \int{{{x}^{2}}{{e}^{-x}}dx} \\\ & =-{{x}^{2}}{{e}^{-x}}+2\int{x{{e}^{-x}}dx} \\\ & =-{{x}^{2}}{{e}^{-x}}+2\left( -x{{e}^{-x}}-{{e}^{-x}} \right) \\\ & =-{{e}^{-x}}\left( {{x}^{2}}+2x+2 \right) \\\ \end{aligned}$$ Now we find the definite from $$\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}=\left[ -{{e}^{-x}}\left( {{x}^{2}}+2x+2 \right) \right]_{o}^{k}=2-{{e}^{-k}}\left( {{k}^{2}}+2k+2 \right)$$ The limit form is $$\displaystyle \lim_{k \to \infty }\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}=\displaystyle \lim_{k \to \infty }\left[ 2-\dfrac{{{k}^{2}}+2k+2}{{{e}^{k}}} \right]$$. We have $$\displaystyle \lim_{k \to \infty }\left[ 2-\dfrac{{{k}^{2}}+2k+2}{{{e}^{k}}} \right]=2-\displaystyle \lim_{k \to \infty }\left[ \dfrac{{{k}^{2}}+2k+2}{{{e}^{k}}} \right]$$. $$\displaystyle \lim_{k \to \infty }\left[ \dfrac{{{k}^{2}}+2k+2}{{{e}^{k}}} \right]=\displaystyle \lim_{k \to \infty }\left[ \dfrac{2k+2}{{{e}^{k}}} \right]=\displaystyle \lim_{k \to \infty }\left[ \dfrac{2}{{{e}^{k}}} \right]=0$$. Therefore, the limit value of $$\displaystyle \lim_{k \to \infty }\int\limits_{0}^{k}{{{x}^{2}}{{e}^{-x}}dx}=\displaystyle \lim_{k \to \infty }\left[ 2-\dfrac{{{k}^{2}}+2k+2}{{{e}^{k}}} \right]=0$$. Therefore, the improper integral converges. **Note:** We have to be careful about the upper boundary of the integral. The choices of functions for the by parts have to be in order, so that the indices value of the ${{x}^{2}}$ reduces. We can’t take them in the opposite order.