convergence

**putnam120** · November 18th 2006, 04:29 PM

Prove that $\lim_{k\to\infty}\int_0^{k}\sin(x)\sin\left(x^2\ri ght)dx$ converges.

**ThePerfectHacker** · November 18th 2006, 06:04 PM

Originally Posted by putnam120

Prove that $\lim_{k\to\infty}\int_0^{k}\sin(x)\sin\left(x^2\ri ght)dx$ converges.

Are you sure it converges?

The standard sine curve,
$\int_0^{\infty} \sin xdx$ diverges by oscillation.

**putnam120** · November 18th 2006, 08:53 PM

yes i am sure that it converges.

here is a hint

$\lim_{k\to\infty}\int_0^k\frac{2x\sin(x)\sin\left( x^2\right)}{2x}dx$

and integration by parts and general applications of limits.

**putnam120** · November 19th 2006, 11:12 AM

here is a solution that my friend gave me.

using the hint we have $\frac{\cos(x^2)\sin(x)}{-2x}\big |_0^{\infty} + \int_0^{\infty}\frac{2x\cos(x)\cos(x^2)}{4x^2}dx-\int_0^{\infty}\frac{\cos(x^2)\sin(x)}{4x^2}dx$

the first expression converges to $-\frac{1}{2}$ . for the third expression we can change the lower limit to 1, since this will not effect the convergence. so we have

$\int_1^{\infty}\frac{\cos(x^2)\cos(x)}{4x^2}dx\le\ int_1^{\infty}\frac{dx}{4x^2}$ which converges.

To handle the second expression we apply integration by parts once again and get.

$\frac{\sin(x^2)\cos(x)}{4x^2}\big |_0^{\infty}-\int_1^{\infty}\frac{4x\sin(x)\sin(x^2)}{16x^4}dx-\int_1^{\infty}\frac{8x\sin(x^2)\cos(x)}{16x^4}dx$

to evaluate the first term we can either use l'hopitals rule or taylor series, your pick, either way you find that it does converge. For the last 2 we use fact that

$\int_1^{\infty}\frac{A\sin(x)\cos(x^2)}{16x^4}dx\l e A\int_1^{\infty}\frac{dx}{16x^4}$ which converges.

**ThePerfectHacker** · November 19th 2006, 01:56 PM

The only mistake is that you cannot write,
$\int_0^{\infty} \frac{2x \sin x\sin x^2}{2x} dx$
You need to change the lower bound because otherwise the functions do not agree everywhere.

Now, I will check what you posted and see if it is correct (I still think that there is no convergence).

**topsquark** · November 19th 2006, 02:15 PM

Originally Posted by ThePerfectHacker

Are you sure it converges?

The standard sine curve,
$\int_0^{\infty} \sin xdx$ diverges by oscillation.

Have you graphed the integrand? For large x the function creates "wave-packets" : oscillations within a sine envelope where, on the inside of the envelope, the wavelength is very small. For all intents the function "fills in" all the empty spaces of the sine envelope, creating a net zero area effect.

I don't see why this integral shouldn't coverge, though obviously my argument above is hardly a proof.

-Dan

**putnam120** · November 19th 2006, 03:17 PM

@Hacker. yes i know. at that point i should have changed the lower limit to 1.

and to topsquark:
if you evaluate it you get $\lim_{k\to\infty}\int_0^{k}\sin(x)dx=\lim_{k\to\in fty}-\cos(x)\big|_0^{k}$
which is just
$\lim_{k\to\infty}-\cos(k)+\cos(0)$ and $\lim_{k\to\infty}\cos(k)$ does not exist because of oscillation.

**ThePerfectHacker** · November 19th 2006, 03:26 PM

Originally Posted by topsquark

.

I don't see why this integral shouldn't coverge, though obviously my argument above is hardly a proof.

What is the limit of $\cos x$ ?

When you approach by $x=2\pi,4\pi,6\pi...$ limit is 1.

When you appraoch by $x=\pi,3\pi,5\pi,...$ limit is -1.

**topsquark** · November 19th 2006, 04:21 PM

Originally Posted by putnam120

@Hacker. yes i know. at that point i should have changed the lower limit to 1.

and to topsquark:
if you evaluate it you get $\lim_{k\to\infty}\int_0^{k}\sin(x)dx=\lim_{k\to\in fty}-\cos(x)\big|_0^{k}$
which is just
$\lim_{k\to\infty}-\cos(k)+\cos(0)$ and $\lim_{k\to\infty}\cos(k)$ does not exist because of oscillation.

Sorry. I didn't look at what I quoted. I was referring to the integrand in the original problem. $sin(x)sin(x^2)$ is the one that oscillates in what I called a "wave packet."

-Dan

**ThePerfectHacker** · November 19th 2006, 04:47 PM

Originally Posted by putnam120

$\int_1^{\infty}\frac{\cos(x^2)\cos(x)}{4x^2}dx\le\ int_1^{\infty}\frac{dx}{4x^2}$ which converges.

Aha! I think I caught a mistake with my trained eye.

If, the continous functions
$0\leq f(x)\leq g(x)$
And, $\smallint g(x)$ converges then so does $\smallint f(x)$ (Of course I mean the improper intergral).

But your inequality does not satisfy the one above, namely because it can be negative.

However, I have a way to fix this error.
Using the integral series test we need to test,
$\sum_{k=1}^{\infty} \frac{\cos k^2 \cos k}{4k^2}$ and see whether it converges. It does since its absolutely convergent, since $|\cos k^2 \cos k|\leq 1$ .

**putnam120** · November 19th 2006, 05:29 PM

$-\int_1^{\infty}\frac{dx}{4x^2}\le\int_1^{\infty}\f rac{\cos(x^2)\cos(x)}{4x^2}dx\le\int_1^{\infty}\fr ac{dx}{4x^2}$
both of which converge to a real value. thus the middle term converges since it is bounded between two convergent series.

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