Show that the integral converges

In summary, The integral $$\int_{0}^{\infty} (sin^2(t) / t^2) dt $$ is convergent. This can be proven by showing that the function ##f(t) = \frac{\sin^2 t}{t^2}## is bounded on the interval [0,1], and then by using the comparison test with the convergent integral ##\int_{1}^{\infty} 1 / t^2 dt##. Additionally, the fact that ##f(t)## approaches 1 as ##t## approaches 0 can also be used as a proof.
  • #1
AllRelative
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2

Homework Statement


(FYI It's from an Real Analysis class.)
Show that $$\int_{0}^{\infty} (sin^2(t) / t^2) dt $$ is convergent.

Homework Equations


I know that for an integral to be convergent, it means that :
$$\lim_{x\to\infty} \int_{0}^{x} (sin^2(t) / t^2) dt$$ is finite.I can also use the fact that let: $$ f(x) = sin^2(t) / t^2 $$
and
Let :$$F(x) = \int_{0}^{x} (sin^2(t) / t^2) dt$$
Since f(x) is always positive from 0 to infinity. If F(x) has an upper limit that is not infinite, than the integral in convergent.

I've also seen a few other concepts around those like absolute convergence, Cauchy's criteria.

The Attempt at a Solution


What I have been able to do is using Chasle relation, I proves it's convergence from 1 to infinity since for values of t from 1 to infinity
$$sin^2(t) / t^2 \leqslant 1 / t^2$$.

And knowing that $$\int_{1}^{\infty} 1 / t^2 dt$$ converges, then we know that $$ \int_{1}^{\infty} (sin^2(t) / t^2) dt$$ converges also.I have no clue on how to attack values of t from 0 to 1.
Thanks for the help.
 
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  • #2
AllRelative said:
I have no clue on how to attack values of t from 0 to 1.
Thanks for the help.

What do you know about ##\frac{\sin t}{t}## for small ##t##?
 
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  • #3
Using Hôpital's rule I can derive the numerator and denominator which gives me (cos t) / 1. So the limit of t approaching 0 is 1.

I am not exactly sure how to use that intuition back to the integral
 
  • #4
AllRelative said:
Using Hôpital's rule I can derive the numerator and denominator which gives me (cos t) / 1. So the limit of t approaching 0 is 1.

I am not exactly sure how to use that intuition back to the integral

What sort of functions are not integrable on ##[0, 1]##?
 
  • #5
AllRelative said:
Using Hôpital's rule I can derive the numerator and denominator which gives me (cos t) / 1. So the limit of t approaching 0 is 1.

I am not exactly sure how to use that intuition back to the integral

Actually, the fact that ##\sin(t)/t \to 1## as ##t \to 0## is the basis of the formulas ##(d/dt) \sin(t) = \cos(t),## etc. Typically, this fact is studied well before anybody mentions l'Hosptal's rule in typical textbooks. After all, unless you know that ##(d/dt) \sin(t) = \cos(t),## you cannot even start to apply l'Hospital's rule!

Another way: since
$$\sin(t) = t - \frac{1}{3!} t^3 + \frac{1}{5!} t^5 - \cdots $$ for all ##t##, we have $$\frac{\sin(t)}{t} = 1 - \frac{1}{3!} t^2 + \frac{1}{5!} t^4 - \cdots $$ for all ##t \neq 0.##
 
  • #6
PeroK said:
What sort of functions are not integrable on ##[0, 1]##?
Functions of the form 1 / x ?
 
  • #7
AllRelative said:
Functions of the form 1 / x ?

More generally. Why does ##1/x## potentially have a problem?
 
  • #8
Well.. When approaches 0 the function explodes to infinity.
 
  • #9
AllRelative said:
I have no clue on how to attack values of t from 0 to 1.
Thanks for the help.

Show that the set of numbers ##\frac{\sin^2 t}{t^2}## for ##t\in (0,1]## has a finite upper bound.

Assume ##\frac{\sin^2 t}{t^2}>2## for some ##t## and find a contradiction with property "##\sin x < x## when ##x>0##".
 
  • #10
AllRelative said:
Well.. When approaches 0 the function explodes to infinity.

The function is unbounded, in other words. What about ##\frac{\sin^2 t}{t^2}##? Is that bounded or unbounded as ##t \rightarrow 0##?
 
  • #11
PeroK said:
The function is unbounded, in other words. What about ##\frac{\sin^2 t}{t^2}##? Is that bounded or unbounded as ##t \rightarrow 0##?
Oh yeah since the value of the function is bounded on the interval [0,1[, the integral converges.

Thanks a lot all of you!
 
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  • #12
AllRelative said:
Well.. When approaches 0 the function explodes to infinity.
Not quite: it matters a lot just "how quickly" the function explodes. For example, ##f(x) = 1/\sqrt{x}## explodes as ##x \to 0## from above, but ##\int dx/\sqrt{x} = 2 \sqrt{x} +C## is OK as ##x \to 0.## The integral ##\int_0^a dx/\sqrt{x} = 2 \sqrt{a}## is valid for all ##a > 0.##
 

1. What does it mean for an integral to converge?

When an integral converges, it means that the area under the curve of the function being integrated is finite, or in other words, the limit of the integral exists and is a real number.

2. How do you determine if an integral converges?

There are several methods for determining if an integral converges, such as the comparison test, the limit comparison test, and the integral test. These methods involve comparing the integral to a known convergent or divergent series or using the properties of integrals to evaluate the limit of the integral.

3. What types of functions typically have convergent integrals?

Functions that are continuous, bounded, and decreasing over their entire domain typically have convergent integrals. This includes most polynomial, exponential, and logarithmic functions.

4. Can an integral converge if the function being integrated is not continuous?

Yes, it is possible for an integral to converge even if the function being integrated is not continuous. As long as the function is bounded and decreases over its entire domain, the integral can still converge.

5. What happens if an integral does not converge?

If an integral does not converge, it means that the limit of the integral does not exist or is infinite. This indicates that the area under the curve of the function being integrated is infinite, or the function is unbounded or oscillates in a way that prevents the integral from converging.

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