5
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ No mistake, $e^{-y}$ definitely 'wins' $\endgroup$ Jan 17, 2015 at 2:58
  • $\begingroup$ Could you help me with why is this so ? $\endgroup$
    – square_one
    Jan 17, 2015 at 2:59
  • 1
    $\begingroup$ Think of L'hopital's rule for $\frac{y^2}{e^y}$... Who wins? $\endgroup$ Jan 17, 2015 at 3:00

4 Answers 4

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for helping me see .. my brain only engages L'Hôpital's rule when it sees limit written explicitly $\endgroup$
    – square_one
    Jan 17, 2015 at 3:04
2
$\begingroup$

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 $\dfrac{\infty}{\infty}$: $$\lim\limits_{t \to \infty}\dfrac{-t^2-2t-2}{e^t}\overset{L}{=}\lim\limits_{t \to \infty}\dfrac{-2t-2}{e^t} \overset{L}{=}\lim\limits_{t \to \infty}\dfrac{-2}{e^t}= 0\text{.}$$

$\endgroup$
1
  • $\begingroup$ the identity you have there is a special case of the Laplace transform :-) $\endgroup$
    – obataku
    Jan 17, 2015 at 7:24
2
$\begingroup$

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 $\int_0^\infty y^2 e^{-y} dy=0+0+2+0+\dots=2$.

$\endgroup$
0
$\begingroup$

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 $.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .