Are there any random variables so that E[X] and E[Y] exist but E[XY] doesn't? Are there any random variables so that E[X] and E[Y] exist but E[XY] doesn't?

Having recently discussed here the difference between "$X$ has expectation" (in the wide sense) and "$X$ is integrable", let us give an example where $X$ and $Y$ are integrable (that is, have finite expectation) but $XY$ does not admit an expectation (that is, ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$).
Let $Z$ be any nonnegative random variable satisfying ${\rm E}(Z)<\infty$ and ${\rm E}(Z^2) = \infty$ (for example, $Z=1/\sqrt{U}$, where $U \sim {\rm uniform}(0,1)$; cf. user8268's comment above). Let $R$ be independent of $Z$ with ${\rm P}(R=1) = {\rm P}(R=-1) = 1/2$. Define the random variables $X$ and $Y$ by $X=ZR$ and $Y=Z$. Then, $X$ and $Y$ have finite expectation. Next note that $XY = Z^2$ if $R=1$ and $XY = -Z^2$ if $R=-1$. Hence,
$$
(XY)^ +  := \max \{ XY,0\}  = Z^2 {\mathbf 1}(R = 1)
$$
and
$$
(XY)^ -   :=  -\min \{ XY,0\} = Z^2 {\mathbf 1}(R = -1),
$$
where ${\mathbf 1}$ denotes the indicator function. Since $Z$ and $R$ are independent (and, by assumption, ${\rm E}(Z^2) = \infty$), we get ${\rm E}(XY)^+ = {\rm E}(XY)^- = \infty$, as desired.
In the example (mentioned above) where $Z=1/\sqrt{U}$, $U \sim {\rm uniform}(0,1)$, one can find that $XY$ has density function $f_{XY}$ given by $f_{XY} = 1/(2x^2)$, $|x|>1$. Thus, $\int_0^\infty  {xf_{XY} (x)\,{\rm d}x}  = \int_1^\infty  {(2x)^{ - 1} \,{\rm d}x}  = \infty $ and $\int_{ - \infty }^0 {|x|f_{XY} (x)\,{\rm d}x}  = \int_{ - \infty }^{ - 1} {(2|x|)^{ - 1} \,{\rm d}x}  = \infty $, corresponding to ${\rm E}(XY)^+ = \infty$ and ${\rm E}(XY)^- = \infty$, respectively.
