I am trying to take the compound distribution of a uniform distribution on the interval $(0, \frac{t}{2})$ where $t$ is parameterized by the exponential distribution $\varepsilon (\lambda)$.
Taking the compound distribution I believe gives the following integral
$$\int_0^\infty \frac 2 t \lambda e^{-\lambda t} \, dt$$
I am not able to evaluate this integral. Have I made a mistake in my reasoning anywhere. Any help would be appreciated.
First of all, notice that: $$P(X<x \mid T = t) = \begin{cases} \displaystyle\int_0^x\frac{2}{t}ds & \text{if } x<\displaystyle\frac{t}{2}\\ 0 & \text{if } x>\displaystyle\frac{t}{2}\\ \end{cases} = \begin{cases} \displaystyle\frac{2x}{t} & \text{if } x<\displaystyle\frac{t}{2}\\ 0 & \text{if } x>\displaystyle\frac{t}{2}\\ \end{cases}$$
Then, using the law of total probability, we get that:
$$P(X<x) = \int_0^{+\infty}P(X<x \mid T = t) f_T(t) \, dt,$$
where $f_T(t) = \lambda e^{-\lambda t}$ is the pdf of $T$. The previous integral is rewritten as follows:
$$P(X<x) = \int_{2x}^{+\infty} \frac{2x}{t} \lambda e^{-\lambda t} \, dt.$$
Unfortunately, the last integral can't be solved using known functions. Moreover, as pointed out by Michael Hardy, I'm not sure that it converges.