Answer
$E[Y] = \infty$
Work Step by Step
$\underline{\text{Verify}\ f_Y(y)\ \text{is a pdf}}$
(a) Note that $y^2\gt 0$ for $y$ positive, so that $y^2\gt 0$ for $y\ge 1\gt 0$. Consequently, $1/y^2\gt 0,\ y\ge 1,$ so that $f_Y(y) \ge 0$ for $y\ge 1$.
(b) Also, $\int_{-\infty}^\infty f_Y(y)\ dy = \int_0^1 1/y^2\ dy = \left. -1/y\right]_1^\infty = -1/\infty - (-1/1) = 0 + 1 =1.$
(a) and (b) together show that $f_Y(y)=1/y^2,\ y\ge 1$ is a valid pdf.
$\underline{\text{Show}\ E[Y] = \infty}$
$\begin{align*}
E[Y] &= \int_{-\infty}^\infty y\cdot f_Y(y)\ dy \\
&= \int_{-\infty}^1 y\cdot f_Y(y)\ dy + \int_1^\infty y\cdot f_Y(y)\ dy \\
&= \int_{-\infty}^1 y\cdot 0\ dy + \int_1^\infty y\cdot \frac{1}{y^2}\ dy \\
&= 0 + \int_1^\infty \frac{1}{y}\ dy \\
&= \left. \ln y\right]_1^\infty \\
&= \ln(\infty) - \ln(1) \\
&= \infty - 0 \\
E[Y] &= \infty.
\end{align*}$
