Bounded functions with unbounded integrals

In summary, the integrator can be shown to be unstable by providing examples of bounded inputs that produce unbounded outputs. Examples for the differentiator, such as ##f(t)=\sin(t^2)##, have already been proven to be unstable. For the integrator, functions like ##f(t)=1## and ##f(t)=u(t)## produce unbounded outputs. Other examples can be found by considering functions that are nonnegative for ##x\ge 0##, bounded, and have an infinite area, such as ##\arctan x,~0\le x##.
  • #1
roam
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Homework Statement



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.

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).##

23h8yzp.jpg


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.
 
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  • #2
roam said:

Homework Statement



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.

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).##

View attachment 139989

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.
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
LCKurtz said:
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]##.

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? :confused:
 
  • #4
roam said:
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? :confused:
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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FAQ: Bounded functions with unbounded integrals

1. What are bounded functions with unbounded integrals?

Bounded functions with unbounded integrals are mathematical functions that have a finite range or values, but their integral (the area under the curve) is infinite. This means that the function may be limited in its output, but it continues to increase or decrease without bound.

2. Can a function be bounded with an unbounded integral?

Yes, a function can be bounded with an unbounded integral. This occurs when the function approaches a limit as the input increases, but the area under the curve continues to increase without bound.

3. What are some examples of bounded functions with unbounded integrals?

One example is the function f(x) = 1/x, which has a range of [0, ∞) but an unbounded integral between 1 and ∞. Another example is the logarithmic function g(x) = ln(x), which is bounded but has an unbounded integral between 1 and ∞.

4. Why are bounded functions with unbounded integrals important?

Bounded functions with unbounded integrals are important in mathematical analysis and real-world applications. They can help us understand the behavior of functions and systems that have finite limits but continue to increase or decrease without bound. They also have practical applications in areas such as physics and economics.

5. How do bounded functions with unbounded integrals impact the convergence of integrals?

Bounded functions with unbounded integrals can have a significant impact on the convergence of integrals. In some cases, the integral may not converge at all, as the function continues to increase or decrease without bound. In other cases, the integral may converge, but the value may be infinite due to the unbounded nature of the function.

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