How can I show that this function is integrable?

Question

Let $r>0$ and $$f \colon \mathbb{R} \to [0,\infty), \,x \mapsto \mathbb{1}_{[0,\infty)}\cdot \exp (-r\cdot x).$$ In order to compute the moments of an exponentially distributed random variable $X$, I would like to show that $p\cdot f$ is integrable wrt. the Lebesgue measure for every polynomial $p$. Intuitionally, this is clear to me, but how can I show it formally?

Answer 1

It suffices to show that $x \mapsto x^k f(x)$ is integrable for $k \geq 1$. Measurability is clear because the functions which we are dealing with are continuous.

Fix some $s \in (0,r)$ and $k \in \mathbb{N}$. If $x \geq 0$, then by the definition of the exponential function

$$\exp(sx) = \sum_{n=0}^{\infty} \frac{(sx)^n}{n!} \geq s^k \frac{x^k}{k!},$$

i.e.

$$x^k \leq s^{-k} k! \exp(sx), \qquad x \geq 0.$$

Consequently,

$$|x^k f(x)| \leq s^{-k} k! \exp(-(r-s)x), \qquad x \geq 0.$$

Since $s$ is strictly smaller than $r$, we have $r-s>0$, and so

$$\int_{[0,\infty)} |x^k f(x)| \, dx \leq s^{-k} k! \int_{[0,\infty)} \exp(-(r-s)x) \, dx < \infty.$$