Proving a theorem about the average value of a function over a specific region

Question

Let's say transient phenomenon in a function. A transient phenomenon is defined as:

"A transient event is a short-lived burst of energy in a system caused by a sudden change of state."

So, for example in the picture below:

The average value of a function over the region $[0,\infty)$ is given by:

$$\overline{y}=\lim_{n\to\infty}\frac{1}{n}\int_0^ny(t)dt\tag1$$

For the example in the picture the average value is equal to $0$.

Theorem: If the function $f(t)$ is the sum of two functions $y(t)$ and $z(t)$. And $y(t)$ is the transient part of the function $f(t)$, the average value of the function $f(t)$ (over the region $[0,\infty)$) is given by:

$$\overline{f}=\lim_{n\to\infty}\frac{1}{n}\int_0^nf(t)dt=\lim_{n\to\infty}\frac{1}{n}\int_0^nz(t)dt\tag2$$

How can I prove that theorem? I've no idea

---

Cheking for a few cases where it does work:

• $$\lim_{n\to\infty}\frac{1}{n}\int_0^n(4+e^{-x}\cos(x))\space dx=\lim_{n\to\infty}\frac{1}{n}\int_0^n4\space dx=4\tag3$$
• $$\lim_{n\to\infty}\frac{1}{n}\int_0^n(\sin^2(t)+e^{-4t}(4+\cos(2t+(\pi/989))))\space dt=$$ $$\lim_{n\to\infty}\frac{1}{n}\int_0^n\sin^2(t)\space dt=\frac{1}{2}\tag4$$

Answer 1

The key here is the limit of $n \rightarrow \infty$. The transient part of the function need not necessarily integrate to 0, as in all your exapmles. However, it must, by definition, approach 0 for $t \rightarrow \infty$ (otherwise it would be in the steady state part of the function). And the factor of $1/n$ suppresses any contribution the transient part has for $n \rightarrow \infty$.

Formally we can show it like this \begin{align} \bar{f} &= \lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n f(t) dt = \lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n \left(y(t) + z(t)\right) dt\\ &= \underbrace{\lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n y(t)dt}_{= 0} + \lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n z(t)dt = \lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n z(t)dt \end{align}

To show that the first term vanishes, we need to investigate $y(t)$ at $t\rightarrow\infty$. Since $\lim_{t\rightarrow\infty}y(t) = 0$, we must have $|y(t)| < t^{-\epsilon} \enspace\forall t > T$ with $\epsilon > 0$, where $T$ is some large time after which $|y(t)|$ strictly decays. For brevity we assume $\epsilon \neq 1$, since we can always replace it with a value smaller than 1, and the inequality still holds. Integrating gives \begin{align} \int_T^n t^{-\epsilon}dt = \left.\frac{t^{1-\epsilon}}{1-\epsilon}\right|_{t=T}^{n} = \frac{n^{1-\epsilon}}{1-\epsilon} - \frac{T^{1-\epsilon}}{1-\epsilon} \end{align} Because we also know that $\left|\int f(t)dt\right| \leq \int |f(t)|dt$, we can write \begin{align} \left|\lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n y(t)dt\right| &\leq \lim_{n\rightarrow\infty}\frac{1}{n}\int_0^n |y(t)|dt = \underbrace{\lim_{n\rightarrow\infty}\frac{1}{n}\overbrace{\int_0^T |y(t)|dt}^{<\infty}}_{=0} + \lim_{n\rightarrow\infty}\frac{1}{n}\int_T^n |y(t)|dt\\ &\leq \lim_{n\rightarrow\infty}\frac{1}{n} \int_T^n t^{-\epsilon}dt = \lim_{n\rightarrow\infty}\frac{1}{n}\frac{n^{1-\epsilon}}{1-\epsilon} - \underbrace{\lim_{n\rightarrow\infty}\frac{1}{n}\overbrace{\frac{T^{1-\epsilon}}{1-\epsilon}}^{<\infty}}_{=0}\\ &= \lim_{n\rightarrow\infty}\frac{n^{-\epsilon}}{1-\epsilon} = 0 \end{align}

We implicitly assumed that $y(t)$ is integrable on all bounded intervals of $t$.