Proving The Average Value of a Function with Infinite Length

Question

This is the given:

One can extend the definition of the average value of a continuous function $f(x)$ to the interval $[a,\infty)$ of infinite length as follows: $$f_{\text{ave}}=\lim_{t\rightarrow\infty}\frac1{t-a}\int_a^tf(x)dx$$

Suppose that $f(x)$ is an arbitrary continuous function. Assume that $f(x) \ge 0$ in the interval $[a,\infty)$ and that the integral $\int_0^{\infty}f(x) \ dx$ is divergent. Prove that $f_{ave}=\lim_{x\rightarrow\infty}f(x)$ if this limit exists.

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My work:

$$f_{\text{ave}}=\lim_{t\rightarrow\infty}\frac1{t-a}\int_a^tf(x)dx = \lim_{t\rightarrow\infty}\frac{F(t)-F(a)}{t-a},~~~ \text{F is an antiderivative of}\ f$$

After this, I'm sort of stuck.

What I'm thinking is: $f(x) \ge 0 \Rightarrow F(x)$ is increasing. So we can continue like so:

$$ = \lim_{t\rightarrow\infty}(f(t)-f(a))$$

by L'Hospital. It seems close but there's still $f(a)$ which leads me to believe that I made a wrong assumption/calculation somewhere. Can someone point me to the right direction?

Answer 1

Certainly the limit $$ \lim_{t \to \infty} \frac{F(t)-F(a)}{t-a} $$ is of type $\frac{\infty}{\infty}$ since $a$ is fixed (so $t-a \to \infty$) and $F(t) = F(0) + \int_0^t f(s)ds$, which diverges as $t \to \infty$, so L'Hospital's Rule applies. Differentiating the top we get $F'(t) = f(t)$ (since derivative of the constant $F(a)$ is 0) and the bottom differentiates to $1$, so the limit converges to $$ \lim_{t \to \infty} \frac{F(t)-F(a)}{t-a} = \lim_{t \to \infty} f(t) $$ as desired.

Answer 2

Let $L=\lim_{x\to\infty}f(x)$ and let $\varepsilon>0$. Choose $M>a$ such that $|f(x)-L|<\varepsilon$ for all $x>M$. If $t>M$, then \begin{align} \left\vert \frac{1}{t-a}\int_a^t f(x)\,dx -L \right\vert &= \left\vert \frac{1}{t-a}\int_a^t (f(x)-L)\,dx \right\vert \\ &\le \left\vert \frac{1}{t-a}\int_a^M (f(x)-L)\,dx + \frac{1}{t-a}\int_M^t (f(x)-L)\,dx \right\vert \\ &\le \frac{M-a}{t-a}\sup_{x\in[a,M]}|f(x)-L| + \frac{t-M}{t-a}\varepsilon. \end{align} It should be easy to see that for sufficiently large $t$, the last expression is less than $2\varepsilon$.