let $X, Y$ be two identically distributed (both are $\mathcal{N}(0,1)$) independent random variables
show that $e^{\frac{X^2}{2}} \in L^1 \iff e^{XY} \in L^1 \iff e^{|XY|} \in L^1$.
my attempt :
1st equivalence :
$$\begin{align} \mathbb{E}[e^{XY}] &= \frac{1}{2\pi}\int_{\mathbb{R}}\int_{\mathbb{R}}e^{xy}e^{-\frac{x^2}{2}}e^{-\frac{y^2}{2}}dydx =\frac{1}{2\pi}\int_{\mathbb{R}}e^{-\frac{x^2}{2}}\int_{\mathbb{R}}e^{xy-\frac{y^2}{2}}dydx \\ &=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-\frac{x^2}{2}}\int_{\mathbb{R}}e^{\frac{x^2}{2}}e^{-\frac{(x-y)^2}{2}}dydx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{\frac{x^2}{2}}e^{-\frac{x^2}{2}}\int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}}e^{-\frac{u^2}{2}}dudx \\ & = \mathbb{E}[e^{\frac{X^2}{2}}] \end{align} $$
I mean yeah this kinda proves that $e^{\frac{X^2}{2}} \in L^1 \iff e^{XY} \in L^1 $
but something is bothering me,
because $\mathbb{E}[e^{\frac{X^2}{2}}] = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}dx = +\infty$
Q1 :
can't we just say that $e^{\frac{X^2}{2}} \in L^1$ is a false claim therefore it can imply anything we desire ?
second equivalence : from the fact that $0< e^{XY} \leq e^{|XY|}$
we conclude that $ e^{|XY|} \in L^1 \implies e^{XY} \in L^1$
$$\begin{align} \mathbb{E}[e^{|XY|}] &= \frac{1}{2\pi}\int_{\mathbb{R}}\int_{\mathbb{R}}e^{|xy|}e^{-\frac{x^2}{2}}e^{-\frac{y^2}{2}}dydx \\ &=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-\frac{x^2}{2}}(\int_{0}^{+\infty}e^{|xy|}e^{-\frac{y^2}{2}}dy +\int_{-\infty}^{0}e^{|xy|}e^{-\frac{y^2}{2}}dy)dx \\ &= \frac{1}{2\pi}[\int_{0}^{+\infty}e^{-\frac{x^2}{2}}(\int_{0}^{+\infty}e^{xy}e^{-\frac{y^2}{2}}dy +\int_{-\infty}^{0}e^{-xy}e^{-\frac{y^2}{2}}dy)dx +\int_{-\infty}^{0}e^{-\frac{x^2}{2}}(\int_{0}^{+\infty}e^{-xy}e^{-\frac{y^2}{2}}dy +\int_{-\infty}^{0}e^{xy}e^{-\frac{y^2}{2}}dy)dx]\\ & \leq \text{Constant}[\mathbb{E}[e^{XY}] + \mathbb{E}[e^{-XY}] ] = \text{Constant}_2[\mathbb{E}[e^{XY}]] \end{align} $$
I used the fact that $-X$ and $Y$ are independant and that $X = -X\, \text{in distribution}$
Q2 :
was my attempt at proving 2nd equivalence correct ?
thanks !
edit 1 : pic of the original problem (it's in french) 
$E[e^{XY}|X]$ exists since $0 \leq e^{XY}$ (it's an extension from $L^2$ to non negative variables!!)
Let's suppose that we don't know the distribution of $X$. And suppose that $E[e^{\frac{X^2}{2}}]<+\infty$
Observe that:
$$\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}(\int_{\mathbb{R}}e^{|xy|}e^{-\frac{1}{2}y^2}dy)dP_X(x)=\frac{2}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{\frac{x^2}{2}}(\int_{]0;+\infty[}e^{-\frac{1}{2}(y-|x|)^2}dy)dP_X(x)=\frac{2}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{\frac{x^2}{2}}(\int_{]-|x|;+\infty[}e^{-\frac{1}{2}u^2}du)dP_X(x) \leq \frac{2}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{\frac{x^2}{2}}(\int_{\mathbb{R}}e^{-\frac{1}{2}u^2}du)dP_X(x) \leq 2E[e^{\frac{X^2}{2}}]<+\infty$$
And then the equivalence!! (the exercise is true if the distribution of $X$ is unknown)