# Homework Help: Prove improper integral exists

1. Dec 26, 2012

### Felafel

1. The problem statement, all variables and given/known data

Prove that
$\int_0^{\infty} sin(x^2)dx$
exists
3. The attempt at a solution

I split the integral in two parts:
$\int_0^1 sin(x^2)dx$ which exists because $sin(x^2)<1$ and so
$\int_0^1 sin(x^2)dx < \int_0^1 1= 1*(1-0)$
but i don't know how to do with the second part:
$\int_0^{\infty} sin(x^2)dx$
i know the maclaurin series converges but don't know how to use this fact

2. Dec 26, 2012

### Staff: Mentor

Your first part needs a lower bound, too (easy to find).

You can find lower and upper bounds for the second part, and I think you can let those bounds converge to the same value.

3. Dec 26, 2012

### Felafel

I have found all the lower bounds, but have some problems with the upper one for the second part.
I thought of having $\int_1^{\infty} sin(x^2) < \int_1^{\infty} 1$ but $\int_1^{\infty} 1$ goes to $\infty$, which wouldn't work.

4. Dec 26, 2012

### Staff: Mentor

Which lower bound did you use for the second part?
At least the one I see gives a natural way to define a similar upper bound.

5. Dec 26, 2012

### Felafel

I thought of:
$\int_1^\infty 1-\frac{1}{x} \leq \int_1^\infty sin(x^2) \leq \int_1^\infty 1+\frac{1}{x}$
so, if $\sum_{k=1}^\infty 1+ \frac{1}{x^n}$converges, my integral exists. but it doesn't converge. (also, i'm not sure my reasoning is correct)

6. Dec 27, 2012

### haruspex

You'll need to make use of the fact that within each complete cycle of the function (e.g. from √(2πn) to √(2πn+2π)) the two halves of the cycle largely, and increasingly, cancel. (If there were a nonzero lower bound on the integral per cycle the integral as a whole would not converge.) See if you can get a diminishing upper bound on the integral over each cycle.

7. Dec 27, 2012

### Ray Vickson

If you change variables to y = x^2 you have that the integral is
$$I = \frac{1}{2} \int_0^{\infty} \frac{\sin(y)}{\sqrt{y}} \, dy,$$
and the integral is the limit as $N \to \infty$ of
$$J(N) = \int_0^N \frac{\sin(y)}{\sqrt{y}} \, dy.$$
If you let $N = n \pi, \; n = 1,2,3, \ldots$ the limit as $n \to \infty$ can be seen to exist (how?). If we let $N \to \infty$ through arbitrary values, we can see that the differences between general $N$ and special $N$ of the form $\pi n$ go to zero for large N.

8. Dec 27, 2012

### Felafel

Thanks everyone. What if I wanted to solve it this way?:

The Maclaurin series for the integral is $sin(x^2)$

$$\sum_{k=1}^{+\infty} \frac{(-1)^k (x^2)^{1+2k}}{(1+2k)!}$$

This sum contains negative and positive value, so i split it in two to have them separately:

Negative values + positive values:
$\sum_{k=1}^{\infty} \frac{(-1)(x^2)^{1+2k}}{(1+2k)!} + \sum_{k=1}^{\infty} \frac{(x^2)^{2k}}{(2k)!}$

i know that: $\sum_{k=1}^{\infty} \frac{x^k}{k!} = e^x$
Thus
$\sum_{k=1}^{+\infty} \frac{(-1)^k (x^2)^{1+2k}}{(1+2k)!} = -e^{2x} + e^2x = 0$

And thus, being the positive and negative parts 0 to the infinity, I can say the series converges.

9. Dec 27, 2012

### Staff: Mentor

I don't see how you get those exponential functions and why you expect them to cancel. The integral is certainly not 0 everywhere.

10. Dec 27, 2012

### Ray Vickson

I do not understand (or believe) your arguments above; in any case, they miss the point completely.

The finite integral can be done using the expansion of sin(t), but your expression is wrong. We have that
$$F(U) \equiv \int_0^U \sin(x^2) \, dx = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} \frac{U^{4k+3}}{4k+3},$$
and this converges for any finite U. However, the issue is whether or not F(U) has a finite limit as U → ∞, and your arguments do not address this problem.

11. Dec 28, 2012

### Felafel

Can i ask you how did you get that expression? because I start to think that the instructions i was given to find the maclaurin expansions are wrong.
Also, how do you find that the summatory has finite limit for k→∞? i keep getting $\frac{\infty}{\infty}$ :(

12. Dec 28, 2012

### Ray Vickson

I got that expression in the standard way: (i) expand sin(t) is a power series in t: $\sin(t) = \sum_{k=0}^{\infty} (-1)^k t^{2k+1}/(2k+1)! .$ (ii) substitute t = x^2, and note that $(x^2)^{2k+1} = x^{4k+2}.$ (iii) integrate term-by term. There are some standard theorems that justify such termwise integration, and there are some standard tests that allow one to say that the result is convergent, etc.

However, you are quite right: using the series expansion is the wrong way to go; I already said that in my previous post. My post was just correcting your mistakes, it was not indicating the right way to do the problem; I had already done that in the first post I made on this subject---did you ever read it?

13. Jan 1, 2013

### Felafel

okay, it took me a while but i think i'm almost there. i really didn't know how to do it with the variations of $pi$ but i changed the variable as you suggested. I simply thought of:

$\int_0^{\infty} \frac{sint}{\sqrt t} \leq \int_0^{\infty} \frac{|sint|}{\sqrt t} \leq \int_0^{\infty} \frac{1}{\sqrt t}$

$\int_0^{\infty} \frac{1}{\sqrt t} = \sum_{t=1}^n \frac{1}{\sqrt t}$ which goes to 0 for $n \to \infty$

I don't think it is formally and correctly written, but do you reckon my reasoning is okay?

14. Jan 1, 2013

### Curious3141

You're missing a $dt$ in all your integrals, but this is correct so far. But not helpful (see below).

This is wrong. Sketch or plot the curve $y = \frac{1}{\sqrt x}$. The discrete summation can be represented by the sum of the areas of rectangles having a constant width and varying height, pencil this in also. What can you say about the comparison between the two areas?

So the relationship is wrong, it's an inequality, and the inequality goes the opposite way from what you need. Not that it matters, because the sum has no limit (just compare with the harmonic sum). And in fact, $\int_0^{\infty} \frac{1}{\sqrt t}dt$ is also unbounded.

15. Jan 1, 2013

### Curious3141

One of the nicest approaches to proving the Fresnel integral converges is to use integration by parts. Start by breaking the bounds up as you did. The first integral between [0,1] is trivial. The second integral between [1, infinity) is the troublesome one. Apply the substitution suggested by Ray Vickson, then integrate by parts.

16. Jan 1, 2013

### Felafel

Oh, okay I didn't realise i messed up with the sequence, i thought it converged.
So, i should have:
$\int_0^1sin(x^2)dx=1+\int_1^{\infty} sin(x^2) dx$
I'll go on with the second part only, substituting y to $x^2$ and integrating by parts:

$=\frac{-cosy}{\sqrt y} |_1^{\infty} - \int_1^{\infty} \frac{cosy}{2y \sqrt y} dy$
from the first part I get:
0+1=1
now i check the integral, which is:

$\int_1^{\infty} \frac{cosy}{2y \sqrt y} dy \leq \frac{1}{2} \sum_{y=1}^{\infty} \frac{1}{2y \sqrt y}$

and the series should now converge.

Or did I mess up again?

17. Jan 1, 2013

### Staff: Mentor

This equation looks wrong.

Remember that you always need and upper and a lower bound - and if they don't approach each other, you need some other way to show convergence. With the transformed integral, this should be easy to do.

18. Jan 1, 2013

### Curious3141

Again, what you wrote here makes no sense. I'll assume those were typos and you meant $\int_0^{\infty}sin(x^2)dx=\int_0^1 sin(x^2)dx+\int_1^{\infty} sin(x^2) dx$

Anyway, you can't say $\int_0^1 sin(x^2)dx = 1$ (because it isn't true). All you can say is $\int_0^1 sin(x^2)dx \leq 1$, which is sufficient for our purposes.
I didn't check the integration by parts too closely, but the form $\int_1^{\infty} \frac{cosy}{2y \sqrt y} dy$ looks to be in the right ballpark. You should end up with something like $\frac{1}{2}\cos 1$ somewhere in there, so recheck your work. But I'm puzzled as to why you're comparing the integral to another discrete summation. Why don't you use similar logic to what you did before, observing that $\cos x \leq |\cos x| \leq 1$ and see how that impacts that definite integral?

19. Jan 1, 2013

### Felafel

yes, i think i've left aside the $\frac{1}{2}$and approximated cos1 to 1.
Anyway, I thought that the improper integral exists when such summations converge.
So should i put it like this intead:

$\int_1^{\infty} \frac{cosy}{2y \sqrt y}dy \leq \int_1^{\infty}\frac{1}{2y \sqrt y}dy= \lim_{n \to \infty} \int_1^n \frac{1}{2y \sqrt y}dy=0$
and say that since the limit is finite the integral exists?

20. Jan 1, 2013

### Curious3141

That's a pretty inaccurate approximation, since cos 1 is very far from 1. Why do you need to approximate this? Just leave it as cos 1.

Correct so far.

And that's wrong, because the definite integral converges to 1.

21. Jan 1, 2013

### Felafel

Lots of thanks for your help!
Just one more (stupid) thing.. why does it converge to 1?

22. Jan 1, 2013

### Curious3141

Work out the indefinite integral and impose the bounds. The upper bound vanishes when you take the limit, leaving you with only the finite lower bound, which evaluates to (-1). 0 -(-1) = 1.

23. Jan 1, 2013

### Ray Vickson

This integration-by-parts is very nice, but what I had in mind originally was a bit different:
$$\int_0^{\pi N} \frac{\sin(t)}{\sqrt{t}}\, dt = a_1 + a_2 + a_3 + \cdots + a_N,\\ a_k = \int_{(k-1) \pi}^{k \pi}\frac{\sin(t)}{\sqrt{t}}\, dt , k = 1,2,\ldots.$$
Because of the sign changes in sin(t) we have
$$a_1 > 0, \, a_2 < 0, \, a_3 > 0, \ldots$$
and $|a_{i+1}| < |a_i|$, with $a_n \to 0$ as $n \to \infty.$ Thus, the series $\sum a_i$ converges, by the alternating-series theorem.

I actually like your integration-by-parts method better.