Finding the moment generating function of the product of two standard normal distributions The following question is on my homework assignment that I cannot figure out:

Let U and V be independent random variables, each having a normal distribution with mean zero and variance one. Find the moment generating function of the random variable W = UV .

I have looked around online, and cannot find an answer to this question. In fact, the only answers I can find that even relate to the product of standard normal random variables are using techniques that we never covered in my class. We covered in class how to find the MGF for linear combinations of random variables, by W isn't linear, its a product of two normals. So that technique won't work. 
What am I supposed to do? I am completely at a loss. I have tried multiplying the MGFs of U and V together, but that leaves me with something ugly that I can't reduce.
 A: The MGF is, by definition, $M(t) = E[e^{tUV}]$.  Try integrating
$$ \dfrac{1}{2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} e^{-v^2/2}\ du\ dv$$
by completing the square.
A: There's a nice way to do this avoiding any tricky integration that I learned from here: observe $XY = (1/4)(X+Y)^2 - (1/4)(X-Y)^2$ and that $X+Y$ and $X-Y$ are independent, so $XY$ is the sum of scalar multiples of $\chi^2$ random variables with one degree of freedom. Using the fact that the MGF of a $\chi^2_1$ random variable is $(1-2t)^{-1/2}$ you can now get the MGF you want.
A: You can use conditioning and representation for a simpler, cleaner approach. You have by conditional expectations that
$$
E\left(e^{tUV}\right) = E\left(E\left(e^{tUV}| U\right)\right) = E\left(e^{\frac{1}{2}t^2V^2}\right)
$$
But notice that $\frac{1}{2}V^2 \sim \frac{1}{2}\chi^2_1 \sim Gamma(1/2)$ in distribution, and so we can use the form of the Gamma MGF.
And so $E\left(e^{tUV}\right) = (1-t^2)^{-\frac{1}{2}}$, which is finite on $t \in (-1,1)$.
