Show that $\int_0^\infty{x^2e^{-x}dx}$ converges without calculating the integral.

Question

So I know how to calculate this, and in doing so showing that it converges: $\int_0^\infty{x^2e^{-x}dx}$. However, I’m not sure how to show that it converges by using series and comparison tests etc.

I figure I’ll have to split it up, $\int_0^\infty{x^2e^{-x}dx} = \int_0^1{x^2e^{-x}dx} + \int_1^\infty{x^2e^{-x}dx}$. And then I suppose the Taylor expansion of $e$ will become important. I’m just not sure where to go from here.

Call the left side $I_0$ and the two parts $I_1$ and $I_2$ respectively, for convenience. I have tried to solve $I_1$ as follows.

\begin{equation} x \rightarrow 0 \Rightarrow e^{x} = 1 + x + \frac{x^2}{2!} + O(x^3) \rightarrow 1 \end{equation} \begin{equation} x \rightarrow 0 \Rightarrow \frac{x^2}{e^{x}} = \frac{x^2}{1} = \frac{1}{x^{-2}} \end{equation}

And so by comparison with $\int_0^1{\frac{1}{x^p}dx}$, $I_1$ converges. Does this work?

I’m not sure how to do $I_2$.

Answer 1

There is no need to split. Compare $x^2e^{-x} $ with $\frac{1}{1+x^2}$ (which is integrable in $[0,+\infty)$) as $x\to +\infty$. We have that $$\lim_{x\to +\infty}\frac{x^2e^{-x}}{1/(1+x^2)}=\lim_{x\to +\infty}\frac{x^2(1+x^2)}{e^{x}}=0$$ which implies that there is $C>0$ such that for $x\geq 0$, $$0\leq x^2e^{-x}\leq \frac{C}{(1+x^2)}.$$

Answer 2

My two cents on this problem: You can show that $$\lim_{x\to\infty} \frac{x^4}{e^x} = 0.$$ Applying the defintion of convergence with $\epsilon = 1$ this means that there is a $x_0 > 0$ such that $$e^{-x} \leq \frac{1}{x^4}, \quad x \geq x_0$$ Now, we can split up the integral als follows: $$\int_{0}^{\infty} x^2 e^{-x} dx = \int_{0}^{x_0} x^2 e^{-x} + \int_{x_0}^{\infty} x^2 e^{-x} \leq \int_{0}^{x_0} x^2 e^{-x} + \int_{x_0}^{\infty} \frac{1}{x^2} < \infty$$ the first integral on the right hand side exists because it's a continous function integrated over a compact interval. The other one is easy to compute.

Answer 3

Without calculating any integral

The function $x^2e^{-x}$ is decreasing if $x>2$. So

$$\int_2^{\infty}x^2e^{-x}dx<\sum_{n=2}^{\infty}n^2e^{-n}.$$

It would be enough then to show that the series on the right hand side is convergent. Let' use the ratio test:

$$\left(\frac{n+1}{n}\right)^2\frac{e^{-(n+1)}}{e^{-n}}=\frac1e\left(\frac{n+1}{n}\right)^2.$$

If $n$ is large enough then the ratio in question will be less than one. That is, our series and our integral are convergent.