Integration by Parts with Probability Density Functions For $Y$ a random variable that has sample space $S = [0, \infty)$ with a density function, $$ f_Y (y) = 9ye^{−3y} $$
Find:
1) $P\{Y > \frac 1 3\}$
2) $\operatorname{E}\left(\frac 1 Y \right)$

I learned how to integrate by parts (years ago, so I'm a bit rusty) and I know there's countless resources online, but in terms of probability density functions I get a bit confused. My notes from lecture uses substitution when simplifying an integral something with gamma or E? I'm guessing this shortcut in simplifying is how to get through this problem.
 A: \begin{align}
\int 9y e^{-3y} \, dy = 9 \int y\Big(e^{-3y} \, dy\Big) = 9 \int y\,dv = 9\left( yv - \int v\,dy \right)
\end{align}
$$
dv = e^{-3y} \, dy, \qquad v = -\frac {e^{-3y}} 3
$$
and so on. In the expression $\left[ yv \vphantom{\dfrac 1 1} \right]_{y=1/3}^{\infty}$ you can use L'Hopital's rule.
A: For the first integral, you have
$$P(Y > \frac{1}{3}) = \int_{1/3}^{\infty} 9ye^{-3y} \> dy$$
Integration by parts is loosely just
$$\int u \> dv = uv - \int v \> du$$
We see that letting $u=y$ will simplify computation. This leaves 
$$P(Y > \frac{1}{3}) = 9 \left( \left. (y(-\frac{1}{3}e^{-3y}))\right|_{1/3}^{\infty} - \int_{1/3}^{\infty} -\frac{1}{3}e^{-3y} \> dy \right)$$
You should be able to evaluate from here. Using LOTUS the expectation is just
$$E(\frac{1}{y}) = \int_0^{\infty} \frac{1}{y} 9ye^{-3y} \> dy = \int_0^{\infty}  9e^{-3y} \> dy$$
A: For 1) compute
$$\int_{1/3}^{\infty} 9ye^{-3y} dy= \lim_{t\to\infty} \int_{1/3}^{t} 9ye^{-3y}dy$$
Using a substitution will make this less tedius, but not by much. To skip that, first take out the $9$, then use $u=y$, $dv=e^{-3y}dy$ in the integration by parts formula:
$$\int_a^b u \ dv = [u(y)v(y)]\Big|_a^b - \int_a^b v \ du$$
For $2$, no parts necessary.
$$E(g(Y)) =\int_0^{\infty} g(y) f_Y(y) dy$$
You may be able to see quickly that the $y's$ cancel for your $g(y)$.
