$\int_0^{\infty}y^2e^{-y} dy$ To calculate $\displaystyle \int_0^{\infty}y^2e^{-y} dy$
=$\displaystyle -y^2e^{-y}-2ye^{-y}-2e^{-y}|_o^{\infty}$
This should fail at $\infty$, but the answer is given as 2. Seems like $e^{-y}$ "wins" over $y^2$. 
So, am I making a mistake here ? Please help.
 A: In that first term, you are evaluating
$$\lim_{y\to \infty}-\frac{y^2}{e^y}=\lim_{y\to\infty}-\frac{2y}{e^y}=\lim_{y\to\infty}-\frac{2}{e^y}=0$$
by repeated application of L'Hopital's rule.
A: A shortcut used on actuarial exams is: (if I recall correctly, for $n > -1$)
$$\int\limits_{0}^{\infty}x^ne^{-ax}\text{ d}x = \dfrac{\Gamma(n+1)}{a^{n+1}}\text{.}$$
So in this case, $$\int\limits_{0}^{\infty}y^2e^{-y}\text{ d}y = \dfrac{\Gamma(3)}{1^{3}} = 2! = 2\text{.}$$
To justify $2$ from where you are at in your question:
$$\begin{align}
\int\limits_{0}^{\infty}y^2e^{-y}\text{ d}y &= \lim\limits_{t \to \infty}\left[-y^2e^{-y}-2ye^{-y}-2e^{-y}\right]^{t}_{0} \\
&= \lim\limits_{t \to \infty}\left[-t^2e^{-t}-2te^{-t}-2e^{-t}\right] - (-0-(-0)-2) \\
&= \lim\limits_{t \to \infty}\left[-t^2e^{-t}-2te^{-t}-2e^{-t}\right] + 2\text{.}
\end{align}$$
So it suffices to show that 
$$\lim\limits_{t \to \infty}\left[-t^2e^{-t}-2te^{-t}-2e^{-t}\right] = \lim\limits_{t \to \infty}e^{-t}\left(-t^2-2t-2\right) = \lim\limits_{t \to \infty}\dfrac{-t^2-2t-2}{e^t} = 0\text{.}$$
Notice that we have an indeterminate form $\dfrac{\infty}{\infty}$ here, so applying L-Hospital, we take derivatives of the numerator and denominator repeatedly until we don't have $\dfrac{\infty}{\infty}$:
$$\lim\limits_{t \to \infty}\dfrac{-t^2-2t-2}{e^t}\overset{L}{=}\lim\limits_{t \to \infty}\dfrac{-2t-2}{e^t} \overset{L}{=}\lim\limits_{t \to \infty}\dfrac{-2}{e^t}= 0\text{.}$$
A: Consider that $\int f(t) e^{-t} dt=-f(t)e^{-t}+\int f'(t) e^{-t}dt$ so $\int_0^\infty f(t) e^{-t} dt=f(0)+f'(0)+f''(0)+\dots$ for functions $f$ such that $f(t)e^{-t}\to0$ as $t\to\infty$. In this case it's then obvious that $\int_0^\infty y^2 e^{-y} dy=0+0+2+0+\dots=2$.
A: When $ y \rightarrow \infty , y^2 \rightarrow \infty \text {  while  } e^{-y}  \rightarrow 0 .$
So , to find the limit of $ y^2e^{-y} \text {   with   } y \rightarrow \infty $ You'd better use L'Hospital rule. I get this limit is 0 , not $\infty $. 
