# Homework Help: Bounded functions with unbounded integrals

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1. Apr 10, 2017

### roam

1. The problem statement, all variables and given/known data

I am trying to show that the integrator is unstable by giving examples of bounded inputs which produce unbounded outputs (i.e. a bounded function whose integral is unbounded).

Note: The integrator is a system which gives an output equal to the anti-derivative of its input.

3. The attempt at a solution

I have already proven the instability of the differentiator by considering the bounded input $f(t)= \sin(t^2),$ which gives the unbounded output $f'(t) = 2t \cos(t^2).$

For the integrator, I know, for instance, that the bounded input $f(t)=1$ gives the unbounded output $t.$ But could anyone suggest a more interesting example like the one I gave for the differentiator?

I couldn't come up with a good example. I would appreciate any suggestions or links.

2. Apr 10, 2017

### LCKurtz

Your problem might be more interesting on a bounded interval. Think about $\sqrt[3] x$ on $(0,1)$ for a differentiation example. But on a bounded interval no example exists for the integration because $\left | \int_a^t f(u)~du\right | \le \int_a^b |f(u)|~du\le M(b-a)$ where $M$ is a bound for $|f|$ on $[a,b]$.

3. Apr 17, 2017

### roam

What about a non-bounded interval? I mean, I already found a function, $f(t)=constant$, which increases indefinitely. The only other example I can think of, would be the Heaviside step function, $f(t)=u(t)$. The output of the integrator for this function is:

$$g(t) = \intop^t_{-\infty} u(\tau) d\tau = \intop^t_{-\infty} 1 \ d\tau = t$$

for $t>0.$

So, is there really no other function whose integral increases indefinitely?

4. Apr 17, 2017

### LCKurtz

Of course not. There are lots of examples. Take any function that is nonnegative for $x\ge 0$ that is bounded but has an infinite area. One such example is $\arctan x,~0\le x$ which is bounded by $\frac \pi 2$.

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