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Math Help - convergence

  1. #1
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    Talking convergence

    Prove that \lim_{k\to\infty}\int_0^{k}\sin(x)\sin\left(x^2\ri  ght)dx converges.
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    Quote Originally Posted by putnam120 View Post
    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.
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  3. #3
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    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.
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  4. #4
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    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.
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  5. #5
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    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).
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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  7. #7
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    @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.
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    Quote Originally Posted by topsquark View Post
    .

    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.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by putnam120 View Post
    @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
    Attached Thumbnails Attached Thumbnails convergence-integrand.jpg  
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  10. #10
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    Quote Originally Posted by putnam120 View Post

    \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.
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  11. #11
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    -\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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