Let X and Y be independent random variables uniformly distributed on (0,a). Find the moment generating function of XY.
I tried to do a double integral from 0 to a and I know that the probability funcion of each X Y is :$\frac{1}{a+1}$: E($e^{txy})= $$\int_0^a\int_0^a e^{txy}\frac{1}{a+1}$$\frac{1}{a+1} dxdy$ . But the integral isnt working well and I thought maybe there is another way to solve it.
Yes, that integral is not going to resolve to elementary functions. You are going to have to call upon the Exponential Integral; a constructed function such that: $$\dfrac{\partial \operatorname {Ei}(cx)}{\partial x}= \dfrac{\exp(cx)}{x}$$
Thus
$$\begin{align}M_{XY}(t)&=\int_0^a\int_0^a \frac {\exp(txy)}{a^2}\operatorname d y\operatorname dx \\[1ex] &= \int_0^a \frac{\exp(atx)-1}{a^2tx}\operatorname d x \\[1ex] & = \left[\dfrac{\operatorname {Ei}(atx)-\ln x}{a^2t}\right]_{x=0}^{x=a} \\[1ex] & = \dfrac{\operatorname {Ei}(a^2t)-\ln a + \raise{0.75ex}{\lim\limits_{x\searrow 0}}(\ln x-\operatorname{Ei}(atx))}{a^2t} \end{align}$$
... assuming said limit is finite.