convergence of $\int^\infty_0 x^2.e^{-ax} dx$

Question

I have to check convergence or divergence of $\int^\infty_0 x^2.e^{-ax} dx$ .

I am following the basic approach of solving such improper integrals and used:

$$\lim_{p\to\infty}\int^p_0x^2.e^{-ax}.dx$$

I have managed to crack it down to:

$$\lim_{b\to\infty}-\left|\frac{x^2.e^{-ax}}{a}+\frac{2xe^{-ax}}{a^2}+\frac{2e^{-ax}}{a^3}\right|^b_0$$

Now, funny thing is. It gives me: $$\lim_{b\to\infty}-e^{-ab}\left[\frac{b^2}{a}+\frac{2b}{a^2}+\frac{2}{a^3}\right]+\frac{2}{a^3}$$

So, basically, it deduces to: $$(\to\infty)\times(\to0)+\frac{2}{a^3}$$

My book says that answer is convergent with integral reducing to $\frac{2}{a^3}$ But I don't see how.

Please help.

Answer 1

First $x^2.e^{-ax}=\mathcal{O}(x^{-2})\implies \int^\infty_0 x^2.e^{-ax} dx\quad$ converges

$\begin{array}{l}\displaystyle \int x^2.e^{-ax} dx&=\displaystyle\left[-x^2\dfrac{e^{-ax}}{a}\right]+\int2x\dfrac{e^{-ax}}{a}\;dx\\&=\displaystyle \displaystyle\left[-x^2\dfrac{e^{-ax}}{a}\right]+\dfrac{2}{a}\left(\left[-x\dfrac{e^{-ax}}{a}\right]+\dfrac{1}{a}\int e^{-ax}\;dx\right)\\ &= \left[-x^2\dfrac{e^{-ax}}{a}\right]-\dfrac{2}{a^2}\left[x\dfrac{e^{-ax}}{a}\right]-\dfrac{2}{a^3}\bigg[e^{-ax}\bigg]\end{array}$

Thus $$\lim_{t\to+\infty}\left[-x^2\dfrac{e^{-ax}}{a}\right]_0^t-\dfrac{2}{a^2}\left[x\dfrac{e^{-ax}}{a}\right]_0^t-\dfrac{2}{a^3}\bigg[e^{-ax}\bigg]_0^t=0-\left(-\dfrac{2}{a^3}\right)=\dfrac{2}{a^3}$$

With $$\color{blue}{\boxed{\lim_{t\to \infty}\dfrac{e^{\alpha t}}{x^{\beta}}=+\infty\quad (\alpha>0)}}$$

Answer 2

Let $p(x)$ be a polynomial of degree $n>0$. Then, $$ \lim_{x\to \infty}\frac{p(x)}{e^x} $$ is indeterminate. And by L'Hopital's apply $n$ times we have $$ \lim_{x\to \infty}\frac{p(x)}{e^x}=\lim_{x\to \infty}\frac{\frac{d^{n-1}}{dx^{n-1}}}{\frac{d^{n-1}}{dx^{n-1}}e^x}=\lim_{x\to \infty}\frac{a}{e^x}=0 $$ where $a$ is the coefficient on the order $n$ term of $p(x)$.

Answer 3

You can evaluate the integral explicitly using Integration by parts just by remembering that $\exp(-x)\rightarrow0\ as\ x\rightarrow\infty$ more fastly then any polynomial going towards infinity, hence it means that $\{x^a*\exp(-x)\rightarrow0\ as\ x\rightarrow \infty\}$ whenever $\text a>0$.

Now the problem just reduced to applying integration by parts thrice and you will get your solution.

Answer 4

$\large (a\ne 0)\quad x^{b}e^{-a x} = e^{-ax+\ln x^b}=e^{u(x)}$

with $\large u(x)= -ax+b\ln x=-ax\left(1-\dfrac{b\ln x}{ax}\right)\sim_{+\infty}-ax$ because $\displaystyle \lim_{x\to +\infty}\dfrac{\ln x}{x}=0 $

Thus $$\Large x^{b}e^{-a x} \underset{+\infty}{\sim}e^{-ax}$$