Integrating $f(y) = e^{-y} y^3$ I can see by checking in Wolfram Alpha that
$$\int_0^{\infty}e^{-y} y^3 dy = 2$$
This leads me to believe that this integral is solvable through standard methods...but I cant see how u-substitution or integration by parts would proceed?
 A: $$
I = \int^{\infty}_{0}\mathrm{e}^{-\alpha y}dy = \frac{1}{\alpha}.
$$
and 
$$
-\frac{d^{3}I}{d\alpha^{3}} = -\frac{d^{3}}{d\alpha^{3}}\int^{\infty}_{0}\mathrm{e}^{-\alpha y}dy = -\int^{\infty}_{0}\frac{d^{3}}{d\alpha^{3}}\mathrm{e}^{-\alpha y}dy\\
=\int^{\infty}_{0}y^{3}\mathrm{e}^{-\alpha y}dy.
$$
Subbing in the for $I$
$$
-\frac{d^{3}I}{d\alpha^{3}}= \frac{6}{\alpha^{4}}
$$
since $\alpha$=1 then the result is 
$$
\int^{\infty}_{0}y^{3}\mathrm{e}^{-y}dy = 6
$$
and not 2.
A: Those some methods


*

*Find the integral using three integrations by parts

*Let 
$$f(n)=\int_0^\infty e^{-y}y^n dy$$
and by one integration by parts find a recursive relation of $f(n+1)$ and $f(n)$

*Use the Gamma function
$$\Gamma(x)=\int_0^\infty e^{-y}y^{x-1} dy$$
and we have
$$\Gamma(n+1)=n!$$

A: An antiderivative of $f(x)=e^{-x}x^2$ is of the form $e^{-y}P(x)$, where $P$ is a degree two polynomial: $P(x)=ax^2+bx+c$; differentiating we get
$$
(e^{-x}P(x))'=-e^{-x}(ax^2+bx+c)+e^{-x}(2ax+b)=e^{-x}(-ax^2+(2a-b)x+b-c)
$$
so we need
\begin{cases}
-a=1\\
2a-b=0\\
b-c=0
\end{cases}
or
$a=-1$, $b=-2$, $c=-2$. Now, computing the integral is easy.
Let's now prove that, if $Q(x)$ is a polynomial, then an antiderivative of $e^{-x}Q(x)$ is of the form $e^{-x}P(x)$ where $P$ is a polynomial having the same degree as $Q$. If the degree of $Q$ is $0$, the statement is obvious. So, let's assume $Q$ has degree $n>0$ and that the statement holds for polynomials having degree less than $n$.
It's not restrictive to assume that $Q(x)=x^{n}+Q_1(x)$, where $Q_1$ has degree at most $n-1$ (the coefficient of $x^n$ can be taken out from the integral).
Thus
$$
\int e^{-x}Q(x)\,dx=
\int e^{-x}x^n\,dx+\int e^{-x}Q_1(x)\,dx=
\int e^{-x}x^n\,dx+e^{-x}P_1(x)
$$
where $P_1$ has degree at most $n-1$, by induction hypothesis. So we have to integrate (by parts)
$$
\int e^{-x}x^n\,dx=-e^{-x}x^n + \int e^{-x}nx^{n-1}\,dx=
e^{-x}x^n+e^{-x}P_2(x)
$$
where $P_2$ has degree (at most) $n-1$. Thus we're done with
$$
P(x)=-x^n+P_2(x)+P_1(x).
$$
(I've left out the constant of integration, which is unimportant as we need one antiderivative.)
