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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.

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    Show the integral you tried to compute. – 2017-02-15
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    PS: the probability density function of a continuous uniform distributed random variable, $X$ over support $(0;a)$ is $f_X(x)=\tfrac 1 a\mathsf 1_{x\in(0;a)}$. – 2017-02-15
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    Have you tried the Taylor series expansion? – 2017-02-15
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    Do you think its better way than this integral? – 2017-02-15
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    Actually, $$E(e^{tXY})= \int_0^a\int_0^a e^{txy}\frac{1}{a^2} dxdy=\int_0^1\int_0^1e^{ta^2xy}dxdy$$ Now, the change of variable $(u,v)=(xy,x)$ yields $dudv=xdxdy=vdxdy$ hence $$E(e^{tXY})=\int_0^1\int_0^ve^{ta^2u}v^{-1}dudv=\int_0^1\int_u^1e^{ta^2u}v^{-1}dvdu=-\int_0^1e^{ta^2u}\ln(u)du$$ which has no expression in terms of elementary functions but has an expression in terms of special functions. – 2017-02-15

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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.

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    Do not be shy! Complete this complement of my comment by showing rigorously that the limit in your last line does exist and leads to the formula $$M_{XY}(t)=g(a^2t)$$ with $$g(t)=\sum_{k=1}^\infty\frac{t^{k-1}}{k\cdot k!}=\frac1t\int_0^t\frac{e^s-1}sds$$ I leave it to you to tell us which member of the exponential integral family this function $g$ corresponds to. – 2017-02-15