• Support PF! Buy your school textbooks, materials and every day products Here!

Show that the integral converges

  • #1

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.
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,463
5,202
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##?
 
  • #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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,463
5,202
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
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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
What sort of functions are not integrable on ##[0, 1]##?
Functions of the form 1 / x ?
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,463
5,202
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
hilbert2
Science Advisor
Insights Author
Gold Member
1,349
419
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,463
5,202
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
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!
 
  • #12
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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.##
 

Related Threads on Show that the integral converges

Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
16
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
977
  • Last Post
Replies
9
Views
527
Top