1. ## convergence

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

2. 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.

3. 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.

4. 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.

5. 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).

6. 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

7. @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.

8. 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.

9. 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

10. 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 $|\cos k^2 \cos k|\leq 1$.

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