If $X$ is a random variable such that $X \geq 0$ a.s., I've been told that $$Xe^{-tX}$$ is an integrable random variable, ($t>0$). But why?
By definition, I should checl that $$E[X e^{-tX}] < \infty$$
Now, the integral is equal to $\int_0^\infty x e^{-tx}f(x)dx$ where $f$ is the density function of $X$.
How can I show it's integrable?
Edit
Since $e^x \geq x$, then $e^{-tx} \leq \frac{1}{tx}$.
Hence the integrand function is bounded in this way: $$x e^{-tx} f(x) \leq \frac{1}{t} f(x)$$
Now I know that $\int_0^\infty t f(x) dx = t \cdot 1$ (as $f$ is a density) and hence $X e^{-tX}$ is an integrable random variable, as it's bounded by an integrable random variable.
Since $e^{x}\geq x$ for $x>0$ you have $e^{-tx}\leq\frac{1}{tx}$, thus $xe^{-tx}\leq \frac{1}{t}$
Can you take it from here?