# Show that the integral converges

AllRelative

## 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.

Homework Helper
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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##?

AllRelative
AllRelative
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

Homework Helper
Gold Member
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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]##?

Homework Helper
Dearly Missed
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.##

AllRelative
What sort of functions are not integrable on ##[0, 1]##?
Functions of the form 1 / x ?

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Functions of the form 1 / x ?

More generally. Why does ##1/x## potentially have a problem?

AllRelative
Well.. When approaches 0 the function explodes to infinity.

Gold Member
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##".

Homework Helper
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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##?

AllRelative
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!

hilbert2