convergence of $\int^\infty_0 x^2.e^{-ax} dx$ 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.
 A: 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 $$\color{blue}{\boxed{\lim_{t\to \infty}\dfrac{e^{\alpha t}}{x^{\beta}}=+\infty\quad (\alpha>0)}}$$
A: 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.
A: Treat $b$ as a variable and look at the terms separately:
$$\lim_{b\to\infty}-e^{-ab}\left(\frac{b^2}a\right) = -\frac1{ae^a}\lim_{b\to\infty}\frac{b^2}{e^b}$$
Now using l'Hôpital's rule, we have
$$\lim_{b\to\infty}\frac{b^2}{e^b} = \lim_{b\to\infty}\frac{2b}{e^b} = \lim_{b\to\infty}\frac2{e^b} = 0$$
A: 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 $p(x)$.
A: $\large (a\ne 0)\quad x^{b}e^{-a x} = e^{-ax+\ln x^b}=e^{u(x)}$ 
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 $$\Large x^{b}e^{-a x} \underset{+\infty}{\sim}e^{-ax}$$
