Is the expected value of this density function $Z$ finite? Is the expected value of this density function $Z$ finite?
$$f_Z(z) =
\begin{cases}
{\dfrac{\ln(z)}{z^2}},  & \text{for $z \ge 1$} \\[2ex]
0, & \text{for $z \lt 1$}
\end{cases}$$
I know expected values can be infinite and when I graph this it looks like it has a horizontal asymptote at $y=0$, so it looks like the expected value can be infinite, but I am not sure? Can someone explain theoretically whether the expected value of this density is finite or infinite?  
 A: The expected value does not exist, or if you prefer is infinite. For the expression for $E(Z)$ is
$$E(Z)=\int_1^\infty z\cdot \frac{\ln z}{z^2}\,dz.$$
For $z\ge e$, we have $\ln z\ge 1$.
Thus by Comparison with $\int_e^\infty \frac{1}{z}\,dz$ our integral diverges.
A: \begin{align}
E(Z)&=\int_{\mathbb R} z f_Z(z) dz\\
&=\int_{-\infty}^\infty z \left[0 \times 1_{z < 1} + \frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\
&=\int_{-\infty}^\infty z [0 \times 1_{z < 1}] dz + \int_{-\infty}^\infty \left[\frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\
&=\int_{-\infty}^\infty \left[\frac{\ln z}{z^2} \times 1_{z \ge 1} dz\right]\\
&=\int_1^\infty z \frac{\ln z}{z^2} dz\\
&=\int_1^\infty \frac{\ln z}{z} dz\\
&= \frac{(\ln z)^2}{2}\biggr\rvert_1^\infty\\
&= \lim_{a \to \infty} \frac{(\ln z)^2}{2}\biggr\rvert_1^a\\
&= \lim_{a \to \infty} \frac{(\ln a)^2}{2}\\
&= \infty\\
\end{align}
or $\displaystyle \lim_{a \to \infty} \frac{(\ln a)^2}{2}$ dne $\displaystyle \because \frac{(\ln a)^2}{2} \to \infty$ as $a \to \infty$
A: Use integration by parts to see that the expected value is infinite. $\int_1^\infty \frac{ln(z)}{z}\,dz=\int_0^\infty u\,du$, where $u=ln(z)$.
