Domain problem when calculating marginal density

Question

I have the following homework assignment: the life expectancy $X$ of a lamp has exponential distribution with rate $\lambda$. The rate depends on the production proccess, such that its population can be described by the uniform distribution $\lambda \sim U[0,1]$. (I'm sorry if the translation is bad, I hope it is understandable).

1. Describe the joint density of the lamp life $X$ and rate $\lambda$.

This one is just $f_{X \mid\lambda}(x\mid\lambda)f_{\lambda}(\lambda) = \lambda e^{-\lambda x}$ for $x \geq 0$ and $\lambda \in [0,1]$ or $0$ elsewhere.

2. Find $X$ distribution.

First I tried to find the marginal density of $X$ by integrating the joint density over $\lambda$: $$ f_{X}(x) = \int_{0}^{1} \lambda e^{-\lambda x} \,d\lambda = -\frac{e^{-x}}{x} - \frac{e^{-x}}{x^{2}} + \frac{1}{x^{2}} \quad \text{if $x \geq 0$ and $0$ otherwise}$$ I am already in trouble, somewhere along the way I lost the ability to compute the marginal density for $X = 0$, which was fine for the joint density, I don't know what happened (the integral is correct) nor do I know why was this information lost. It gets even uglier when I try to find the distribution of $X$: $$F(X \leq x) = \int_{0}^{x} -\frac{e^{-x}}{x} - \frac{e^{-x}}{x^{2}} + \frac{1}{x^{2}} \,dx = (-\frac{1}{x} + \frac{e^{-x}}{x})\Big|_{0}^{x} $$ Once again the integral is correct but now I can't even evaluate it because of the lower limit being $0$. So my question is what am I doing wrong? And why did I lose the information on $X = 0$? I know I can just calculate $F(X \leq x \mid \lambda)$ and get some nice results after integrating over $\lambda$ but I really don't know why this straight forward way doesn't really work here...

Thank you for your attention.

Answer 1

The marginal density \begin{align}f(x)&=\int_0^1 \lambda e^{-\lambda x} \text{d}\lambda\\ &= \int_0^1 -\frac{\partial}{\partial x} e^{-\lambda x} \text{d}\lambda\\ &= -\frac{\partial}{\partial x}\int_0^1 e^{-\lambda x} \text{d}\lambda\\ &= -\frac{\partial}{\partial x} \frac{1-e^{-x}}{x}\tag{1}\\ &= \frac{1-e^{-x}}{x^2} -\frac{e^{-x}}{x}\end{align} is well-defined over $(0,\infty)$. The fact that it goes to infinity at $x=0$ does not matter since a density is uniquely defined almost everywhere. Setting $f(0)=0$ or $f(0)=12346$ is equally valid (and irrelevant). The important fact is that $f$ is integrable over $(0,\infty)$, with mass equal to one.

The cdf is found in (1): since $$\lim_{x\to 0} \frac{1-e^{-x}}{x} = \lim_{x\to 0} \frac{\frac{\partial}{\partial x} 1-e^{-x}}{\frac{\partial}{\partial x} x} = \lim_{x\to 0} \frac{e^{-x}}{1} = 1 $$ by L'Hospital's rule, $$F(x)= 1-\frac{1-e^{-x}}{x}$$