5
$\begingroup$

I've come across a rather difficult expression to integrate. $$\int^\infty_0\frac{x^2e^{-x/y}}y\,dx$$ Is there an easy way to solve this? I've solved it manually which involved multiple u-subs and integration by parts, and took up a decent amount of my time. This comes from a probability textbook so I doubt they want me to spend so much time solving an integral.

The final answer should be $2y^2$.

$\endgroup$
  • 1
    $\begingroup$ The book probably assume you know the identity $\int_0^\infty t^n e^{-t} dt = n!$. This integral representation of $n!$ is useful when you want to transform series which contains $n!$ in its numerator. $\endgroup$ – achille hui Dec 11 '17 at 23:53
9
$\begingroup$

Let us define $z = x/y$, so $dz=dx/y$. Hence, we need to compute the following integration: $$y^2 \int_0^{\infty} z^2 e^{-z} dz = y^2 \Gamma(3)=2y^2.$$

The definition of the $\Gamma$ function: $$\Gamma (n) = \int_0^{\infty} z^{n-1} e^{-z} dz,$$ and $$\Gamma (n) = (n-1) \Gamma(n-1)=(n-1)!.$$

$\endgroup$
  • $\begingroup$ I guess I have to become more used to recognizing when the gamma function is appropriate, thanks $\endgroup$ – Allan Dec 12 '17 at 0:08
7
$\begingroup$

Since $y$ is constant, a change of variables $u = x/y$ means that you just need to compute $$y^2 \int_0^{\infty} u^2 e^{-u}\, du$$ which is handled by two applications of integration by parts.

$\endgroup$
3
$\begingroup$

Using a probabilistic approach denote $X\sim \mathcal{E}xp (1/y)$, hence $$ Var(X) = \mathbb{E}X^2 - \mathbb{E}^2X = y^2, $$ hence, $$ \mathbb{E}X^2 = \int_0^{\infty}x^2\frac{1}{y}e^{-x/y}dx=Var(X)+\mathbb{E}^2X=y^2+y^2=2y^2. $$ Basically, without probability you can just use integration by parts by denoting $u'_x = \frac{1}{y}e^{-x/y}$ and $v=x^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.