We are given two independent standard normal random variables X and Y. We need to find out the M.G.F of XY.
Let $u=XY$
$M_{XY}(t)=M_{u}(t)=E(e^{xyt}) =E(e^{ut})= \int_{u=-\infty}^{+\infty}e^{ut} f_u(u) du, u =xy $
Now how to split $f_{xy}(xy)$
I saw the solution in Finding the M.G.F of product of two random variables.
i recently started probability and statistics. I did not understand how $f_{XY}(XY)$ is split (here it is actually not joint distribution of 2 random variables right? it is pdf of single random variable XY). Kindly elaborate
First, don't switch between cases. Mathematics' symbols are case sensitive.
Further, it is preferred standard to use upper-case for random variables, and lower-case for scalar terms (constants, parameters, et cetera).
Now, by the definition, a moment generating function of random variable $U$ with parameter $t$ is the expectation of random variable $\mathrm e^{tU}$: $$\mathsf M_U(t)=\mathsf E(\mathrm e^{tU})$$
When $U$ is defined as the product of random variables $X$ and $Y$, we therefore have: $$\mathsf M_{XY}(t)=\mathsf E(\mathrm e^{tXY})$$
By the Law of Total Expectation, and the fact that the random variables are independent.
$$\begin{align}\mathsf M_{XY}(t)&=\mathsf E(\mathsf E(\mathrm e^{(tX)Y}\mid X))\\&=(2\pi)^{-1}\int_\Bbb R \mathrm e^{-x^2/2}\int_\Bbb R \mathrm e^{-y^2/2}\cdot\mathrm e^{txy}~\mathrm d y~\mathrm d x\\&=\sqrt{2\pi~}^{-1}\int_\Bbb R \mathrm e^{-x^2/2}\cdot \mathrm e^{t^2x^2/2}~\mathrm d x\\&=\sqrt{2\pi~}^{-1}\int_\Bbb R\mathrm e^{-(1-t^2)x^2/2}~\mathrm d x\\&=\sqrt{1-t^2~~}^{-1}\quad\big[\Re(t^2)<1\big] \end{align}$$
Taking that a few smaller steps at a time.
$$\mathsf M_{XY}(t)=\mathsf E(\mathsf E(\mathrm e^{(tX)Y}\mid X))\\=\mathsf E(\mathsf M_Y(tX))$$
Now, $Y\sim\mathcal{N}(0,1^2)$ (standard normal), so $\mathsf M_Y(s)=\mathrm e^{s^2/2}$. If you do not already have that available$$\begin{align}\mathsf M_Y(s)&=\mathsf E(\mathrm e^{sY})\\&=\int_\Bbb R\mathrm e^{sy}\cdot\sqrt{2\pi~}^{-1}\mathrm e^{-y^2/2}\mathrm d y\\&=\sqrt{2\pi}^{-1}\int_\Bbb R\mathrm e^{sy-y^2/2}\mathrm d y\\&=\mathrm e^{s^2/2}\end{align}$$
... Okay. Anyway, using that:
$$\begin{align}\mathsf M_{XY}(t)&=\mathsf E(\mathrm e^{(tX)^2/2})\\&=\int_\Bbb R\mathrm e^{t^2x^2/2}\cdot\sqrt{ 2\pi~}^{-1}~\mathrm e^{-x^2/2}~\mathrm d x\\&=\sqrt{ 2\pi~}^{-1}\int_\Bbb R\mathrm e^{-(1-t^2)x^2/2}\mathrm d x\\&=\sqrt{1-t^2~}^{-1}\quad\big[\Re(t^2)<1\big]\end{align}$$