just by looking at the graph of the density function can you determine if the expected value is finite or infinite? for example can you conclude that the expected value of $Z$ is infinite just by looking at the graph?  
$$f_Z(z) =
\begin{cases}
{\dfrac{\ln(z)}{z^2}},  & \text{for $z \ge 1$} \\[2ex]
0, & \text{for $z \lt 1$}
\end{cases}$$
here is the graph
 A: Looking by the graph is never sufficient. Pick $a < -1$. Then
$$
\int_1^{\infty} x^a \, dx = \left. \frac{x^{a+1}}{a+1} \right|_1^{\infty} = \frac {-1}{a+1}
$$
(note that $a < -1$ means $a+1 < 0$, so that the above fraction denotes a positive area under a positive curve) but for $a=-1$, 
$$
\int_1^{\infty} \frac 1x \, dx = \left. \phantom{\int} \hspace{-12 pt} \log(x) \right|_1^{\infty} = \infty. 
$$
You could argue that for $a$ close to $-1$, a finite portion of the graph of $x^a$ will look extremely similar to the graph of $x^{-1}$, so you cannot come to any relevant conclusions with just a picture.
In your case, letting $u = \log x$, $du = \frac 1x dx$, we have
$$
\int_1^{\infty} x \frac{\log x}{x^2} \, dx = \int_1^{\infty} \frac{\log x}x \, dx = \int_0^{\infty} u \, du = \left. \frac{u^2}2 \right|_0^{\infty} = \infty. 
$$
Hope that helps,
A: Although not an answer to the OP, I feel like the expectation of the example should be computed.
$$
\mathbb{E}(Z)=\int_1^\infty z\cdot f_Z(z)dz=\int_1^\infty z\cdot\frac{\ln(z)}{z^2}dz=\int_1^\infty\frac{\ln(z)}{z}dz
$$
With a $u$-substitution, where $u=\ln(z)$ and $du=\frac{1}{z}$, this equals
$$
\int_0^\infty udu
$$
which diverges.
