# Show that the integral converges

## 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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PeroK
Homework Helper
Gold Member
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
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

PeroK
Homework Helper
Gold Member
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 $[0, 1]$?

Ray Vickson
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 $t \neq 0.$

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

PeroK
Homework Helper
Gold Member
Functions of the form 1 / x ?
More generally. Why does $1/x$ potentially have a problem?

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

hilbert2
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 $x>0$".

PeroK
Homework Helper
Gold Member
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 $t \rightarrow 0$?

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
Ray Vickson
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 $a > 0.$