Prove that $\displaystyle \lim_{k\to\infty}\int_0^{k}\sin(x)\sin\left(x^2\ri ght)dx$ converges.
here is a solution that my friend gave me.
using the hint we have $\displaystyle \frac{\cos(x^2)\sin(x)}{-2x}\big |_0^{\infty} + \int_0^{\infty}\frac{2x\cos(x)\cos(x^2)}{4x^2}dx-\int_0^{\infty}\frac{\cos(x^2)\sin(x)}{4x^2}dx$
the first expression converges to $\displaystyle -\frac{1}{2}$. for the third expression we can change the lower limit to 1, since this will not effect the convergence. so we have
$\displaystyle \int_1^{\infty}\frac{\cos(x^2)\cos(x)}{4x^2}dx\le\ int_1^{\infty}\frac{dx}{4x^2}$ which converges.
To handle the second expression we apply integration by parts once again and get.
$\displaystyle \frac{\sin(x^2)\cos(x)}{4x^2}\big |_0^{\infty}-\int_1^{\infty}\frac{4x\sin(x)\sin(x^2)}{16x^4}dx-\int_1^{\infty}\frac{8x\sin(x^2)\cos(x)}{16x^4}dx$
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
$\displaystyle \int_1^{\infty}\frac{A\sin(x)\cos(x^2)}{16x^4}dx\l e A\int_1^{\infty}\frac{dx}{16x^4}$ which converges.
The only mistake is that you cannot write,
$\displaystyle \int_0^{\infty} \frac{2x \sin x\sin x^2}{2x} dx$
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).
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
@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 $\displaystyle \lim_{k\to\infty}\int_0^{k}\sin(x)dx=\lim_{k\to\in fty}-\cos(x)\big|_0^{k}$
which is just
$\displaystyle \lim_{k\to\infty}-\cos(k)+\cos(0)$ and $\displaystyle \lim_{k\to\infty}\cos(k)$ does not exist because of oscillation.
Aha! I think I caught a mistake with my trained eye.
If, the continous functions
$\displaystyle 0\leq f(x)\leq g(x)$
And, $\displaystyle \smallint g(x)$ converges then so does $\displaystyle \smallint f(x)$ (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,
$\displaystyle \sum_{k=1}^{\infty} \frac{\cos k^2 \cos k}{4k^2}$ and see whether it converges. It does since its absolutely convergent, since $\displaystyle |\cos k^2 \cos k|\leq 1$.
$\displaystyle -\int_1^{\infty}\frac{dx}{4x^2}\le\int_1^{\infty}\f rac{\cos(x^2)\cos(x)}{4x^2}dx\le\int_1^{\infty}\fr ac{dx}{4x^2}$
both of which converge to a real value. thus the middle term converges since it is bounded between two convergent series.