Average value of $e^{-x}$ from $0$ to $\infty$

Question

I just signed up for this site so I hope I'm doing this right.

I've been having trouble wrapping my head around the average value of the function $e^{-x}$ from 0 to $\infty$. I know that the average value of a function is:

$$\bar f =\frac{\int_a^b{f(x)dx}}{\int_a^b{dx}}=\frac{\int_a^b{f(x)dx}}{b-a}$$

And when a is 0 and b is $\infty$:

$$\frac{\int_0^\infty{f(x)dx}}{\int_0^\infty{dx}}$$

And taking the limit since the integral is improper:

$$\lim_{b\to\infty}\frac{\int_0^b{f(x)dx}}{\int_0^b{dx}}$$

When I evaluate the average value of the function $e^{-x}$ I get a result of zero (one over infinity) by taking the following steps:

$$\lim_{b\to\infty}\frac{\int_0^b{e^{-x}dx}}{\int_0^b{dx}}=\lim_{b\to\infty}\frac{-e^{-x}|_0^b}{x|_0^b}=\lim_{b\to\infty}\frac{-e^{-b}+1}{b}=0$$

While this makes some mathematical sense to me (Unless I made a mistake in which case please let me know!) I'm puzzled on how the average value of this function can be zero when all defined values of the same function are greater than zero. Does this conclusion make any sense or am I off-track entirely?

Answer 1

Consider the option that the "average" is not zero. Then I guess we can agree that it must be positive. However, for every positive number $\varepsilon>0$ we can choose a large enough $b$ such that the average on $[0,b]$ is less than $\varepsilon$. Therefore, the average has to be zero.

The thing is, when using infinity one can often find some results that are not immedietely intuitive.

Answer 2

I find it intuitive. We start with a value of $1=f(0)$ and decrease to $0$ as $n \to \infty$. If we find $f(n)$ for $n=0,1,2,3,...$ The more terms we choose, the more terms we have close to zero, and thus our average gets closer to $0$. Just think about it if you have $100$ numbers close to $0$ despite that $1$ in the begging isn't our average going to be close to $0$?. Although what I'm describing is more of an approximation with rectangles (width $=1$ and thus its height $f(n)$ is the area), the same idea applies if you use $f(n)$ $n=0,0+h,0+2h,...$

Answer 3

The definition of average value as $\overline f =\dfrac{\int_a^b{f(x)dx}}{\int_a^b{dx}}=\frac{1}{b-a}\int_a^b{f(x)dx}$ makes a key assumption: Every value of $x$ in $[a,b]$ is equally likely. For finite intervals, this is a perfectly acceptable model. But it is not the only one possible, however, and for infinite intervals it is not an option at all ($x$ would have zero probability of being found in any finite interval.)

The above is formalized by saying that the probability distribution function $p(x)$ of $x$ is $p(x)=\frac{1}{b-a}$ for $x\in[a,b]$ and zero otherwise. In that case, the original definition is equivalent to $\bar f =\int f(x) p(x)dx$ and it is this form which generalizes. As an example, one could consider the case of $p(x)=e^{-x}$ for $x\geq 0$. Under this assumption, the average value of $x^n$ for integer $n$ is $\overline{x^n}=\int_0^\infty x^n e^{-x}\,dx=n!$. Many other cases are possible, but average values of functions are defined in the same way.