Why does the exponential distribution's pdf integrate to 1?

Question

From All of Statistics pg. 29:

EXPONENTIAL DISTRIBUTION. $X$ has an Exponential distribution with paramater $\beta$, denoted by $X \sim \text{Exp}(\beta)$, if

$$ f(x) = \frac{1}{\beta}e^{-x/\beta} \text{s.t. } x > 0 $$

where $\beta > 0$. The exponential distribution is used to model the lifetimes of electronic components and the waiting times between rare events.

But this must mean (in order for the probability density function to make sense) that

$$ 1/\beta \int_{0}^{\infty} e^{-x / \beta} = 1 $$

so that

$$ \int_{0}^{\infty} e^{-x / \beta} = \beta $$

But why is this?

Answer 1

We have that $$\frac{d}{dx}e^{-\frac{1}{\beta}x}=-\frac{1}{\beta}e^{-\frac{1}{\beta}x}.$$

So that $$\int_0^\infty\frac{1}{\beta}e^{-\frac{1}{\beta}x}=\lim_{t\rightarrow\infty}\left[-e^{-\frac{1}{\beta}x}\right]_0^t =\lim_{t\rightarrow\infty}\left(-e^{-\frac{1}{\beta}t}+1\right)=1.$$

This is an improper integral, which is why the limit procedure is necessary. The exponential decays to zero as $t\rightarrow\infty$ since $\beta>0.$

Answer 2

Indeed, $$\int_0^{\infty}e^{-x/\beta}\;dx=\int_0^{\infty}d\left(-\beta e^{-x/\beta}\right)=\left.-\beta e^{-x/\beta}\right|_0^{\infty}$$ $$= \lim_{R\to\infty}\left.-\beta e^{-x/\beta}\right|_0^R=\lim_{R\to\infty}(-\beta e^{-R/\beta}-(-\beta))$$ $$=0+\beta = \beta$$