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Math Help - Integral Convergence Theorem

  1. #1
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    Lightbulb Integral Convergence Theorem





    I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?

    But I'm still not sure how we can solve this.

    Any help would be much appreciated!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post




    I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?

    But I'm still not sure how we can solve this.

    Any help would be much appreciated!
    How rigorous are you trying to be? Letting z=nx\implies dz=ndx\implies \frac{dz}{n}. So, our integral becomes \frac{1}{n}\int_0^{\pi n}\frac{\sin(z)}{z}dz and thus \lim_{n\to\infty}\frac{1}{n}\int_0^{\pi n}\frac{\sin(z)}{z}dz=\lim_{n\to\infty}\frac{1}{n}  \cdot\lim_{n\to\infty}\int_0^{\pi n}\frac{\sin(z)}{z}dz=0\cdot\frac{\pi}{2}=0

    If you don't know or can't use the fact that \int_0^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2} you can use the following technique.

    Let f(n)=\int_0^{\pi n}\frac{\sin(x)}{x}dx. Then, f(n+1)-f(n)=\int_{\pi n}^{\pi(n+1)}\frac{\sin(x)}{x}dx but on [\pi,(n+1)\pi],\text{ }n\in\mathbb{N} the sine function is negative and \frac{1}{x} positive and thus f(n+1)\leqslant f(n).

    It follows that \lim_{n\to\infty}\frac{1}{n}\int_0^{\pi n}\frac{\sin(x)}{x}dx\leqslant\lim_{n\to\infty}\fr  ac{1}{n}\int_0^{\pi}\frac{\sin(x)}{x}dx=0 and so the conclusion follows if you can prove...
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  3. #3
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    Thanks.

    If we instead break the integral into two parts [0,ε], [ε,pi], can we interchange the lim and integral signs to evaluate?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Thanks.

    If we instead break the integral into two parts [0,ε], [ε,pi], can we interchange the lim and integral signs to evaluate?
    That above theorem, isn't a\in[a,b]

    Also, do you know the DCT?
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  5. #5
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    Let me see if I know enough to solve this...

    So if you do want to integrate this by DCT:

    So I know that  | \frac { \sin nx }{ nx } | \leq \frac {1}{nx} for  x \in (0, \pi )

    Therefore we can place limit inside the integral, so it is:

     \int ^ \pi _0 \lim _{n \rightarrow \infty } \frac { \sin nx }{ nx } dx Is this right so far?

    Well, then, doesn't it converge to 0?
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Let f(n)=\int_0^{\pi n}\frac{\sin(x)}{x}dx. Then, f(n+1)-f(n)=\int_{\pi n}^{\pi(n+1)}\frac{\sin(x)}{x}dx but on [n\pi,(n+1)\pi],\text{ }n\in\mathbb{N} the sine function is negative only for odd n and \frac{1}{x} positive and thus f(n+1)\leqslant f(n).
    There is a mistake here. However, the first proof was fine (by the way, it doesn't require the value of the integral \int_0^\infty\frac{\sin x}{x}dx, just its improper convergence)

    Quote Originally Posted by kingwinner View Post
    I think it is related to the integral convergence theorem, but not all conditions are satisfied. To use the theorem, maybe we have to integrate over something like [ε,pi]?
    You can indeed procede this way. For x>\varepsilon, for all n\geq 1, \frac{|\sin (nx)|}{nx}\leq \frac{1}{n\varepsilon}, which implies that the sequence f_n:x\mapsto \frac{\sin (nx)}{nx} converges uniformly to 0 on [\varepsilon,\pi] (but not on (0,\pi] since the limit at 0 is 1 hence the limit is discontinuous). By your theorem, the integral on [\varepsilon,\pi] converges to 0. As for the piece on [0,\varepsilon], you can prove its absolute value is less than \varepsilon.


    Quote Originally Posted by tttcomrader View Post
    Let me see if I know enough to solve this...

    So if you do want to integrate this by DCT:

    So I know that  | \frac { \sin nx }{ nx } | \leq \frac {1}{nx} for  x \in (0, \pi )
    You should read the statement of the dominated convergence theorem again: you need to dominate the sequence of functions by a function that does not depend on n anymore and that is integrable. Neither of these constraints is fulfilled by your bound. What would work is simply use | \sin (nx)|\leq nx to prove \left|\frac{\sin (nx)}{nx}\right|\leq 1, since 1 is integrable on [0,\pi]. This proof using the TCD is much shorter, but much less elementary than the other ones.
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  7. #7
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    Smile

    Quote Originally Posted by Laurent View Post
    You can indeed procede this way. For x>\varepsilon, for all n\geq 1, \frac{|\sin (nx)|}{nx}\leq \frac{1}{n\varepsilon}, which implies that the sequence f_n:x\mapsto \frac{\sin (nx)}{nx} converges uniformly to 0 on [\varepsilon,\pi] (but not on (0,\pi] since the limit at 0 is 1 hence the limit is discontinuous). By your theorem, the integral on [\varepsilon,\pi] converges to 0. As for the piece on [0,\varepsilon], you can prove its absolute value is less than \varepsilon.
    Hi Laurent,

    "...but not on (0,\pi] since the limit at 0 is 1 hence the limit is discontinuous..."
    Why at x=0, the limit as n->infinity is 1? I think sin(nx)/(nx) isn't even defined when x=0? Can you explain in more detail why it is NOT uniformly convergent on (0,pi], please?

    Also, I don't understand how to handle the [0,ε] piece. Can you explain a little more?

    Your help is much appreciated!
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  8. #8
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    Quote Originally Posted by kingwinner View Post
    Hi Laurent,

    "...but not on (0,\pi] since the limit at 0 is 1 hence the limit is discontinuous..."
    Why at x=0, the limit as n->infinity is 1? I think sin(nx)/(nx) isn't even defined when x=0? Can you explain in more detail why it is NOT uniformly convergent on (0,pi], please?
    Ok, I assumed you let the function equal 1 at 0 since this is the limit (remember \lim_{x\to0}\frac{\sin x}{x}=1), but you don't have to.

    Suppose the convergence were uniform on (0,\pi]. For every n, you have \lim_{x\to 0^+} f_n(x)=1 like I said. However, you can prove that this, together with the uniform convergence, implies \lim_{x\to 0^+}f(x)=1, where f is the limit of f_n (write |f(x)-1|\leq |f_n(x)-f(x)|+|f_n(x)-1|<2\epsilon for suitable choices of large n and small x...). In the present case, f(x)=0 for 0<x<\pi, hence clearly the previous limit is wrong: this is a contradiction.


    Also, I don't understand how to handle the [0,ε] piece. Can you explain a little more?
    Just use \left|\frac{\sin nx}{nx}\right|\leq 1 inside the integral.
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  9. #9
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    Quote Originally Posted by Laurent View Post
    Ok, I assumed you let the function equal 1 at 0 since this is the limit (remember \lim_{x\to0}\frac{\sin x}{x}=1), but you don't have to.

    Suppose the convergence were uniform on (0,\pi]. For every n, you have \lim_{x\to 0^+} f_n(x)=1 like I said. However, you can prove that this, together with the uniform convergence, implies \lim_{x\to 0^+}f(x)=1, where f is the limit of f_n (write |f(x)-1|\leq |f_n(x)-f(x)|+|f_n(x)-1|<2\epsilon for suitable choices of large n and small x...). In the present case, f(x)=0 for 0<x<\pi, hence clearly the previous limit is wrong: this is a contradiction.
    OK, but the definition given in my textbook is this...
    To show f_n(x) does not converge uniformly to 0 on (0,pi], we need to show that
    sup| f_n(x)-0| does NOT converge to 0 as n->infinity.
    0<x≤pi
    How can we prove using this definition?

    Just use \left|\frac{\sin nx}{nx}\right|\leq 1 inside the integral.
    Is this always ≤1?
    |sin(nx)/(nx)| 1/nx What if x is very small?

    Also, I don't see how this is going to help us to evaluate
    ε
    ∫ sin(nx)/(nx) dx
    0
    What is the connection?

    I hope someone can explain this. Thank you!
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  10. #10
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    Quote Originally Posted by kingwinner View Post
    OK, but the definition given in my textbook is this...
    To show f_n(x) does not converge uniformly to 0 on (0,pi], we need to show that
    sup| f_n(x)-0| does NOT converge to 0 as n->infinity.
    0<x≤pi
    How can we prove using this definition?
    Here, we have \sup |f_n(x)-0|\geq 1 because 1 is the limit when x\to 0^+, hence this doesn't converge to 0.

    Is this always ≤1?
    yes, |\sin x|\leq x for all x\geq 0. For instance, because |\sin x|=\left|\int_0^x -\cos t dt\right|\leq \int_0^x |\cos t|dt\leq \int_0^x dt=x.

    Also, I don't see how this is going to help us to evaluate
    ε
    ∫ sin(nx)/(nx) dx
    0
    What is the connection?
    The point is not to evaluate the integral (you can't) but to prove it is small when \epsilon is small.

    The whole scheme is like: let \epsilon>0. Using the theorem, we have \int_\epsilon^\pi f_n(x)dx\to_n 0 hence for n\geq n_0, \left|\int_\epsilon^\pi f_n(x)dx\right|\leq \epsilon. Furthermore, \left|\int_0^\epsilon f_n(x)dx\right|\leq \epsilon. Thus, \left|\int_0^\pi f_n(x)dx\right|\leq 2\epsilon.
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  11. #11
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    Quote Originally Posted by Laurent View Post
    Here, we have \sup |f_n(x)-0|\geq 1 because 1 is the limit when x\to 0^+, hence this doesn't converge to 0.


    yes, |\sin x|\leq x for all x\geq 0. For instance, because |\sin x|=\left|\int_0^x -\cos t dt\right|\leq \int_0^x |\cos t|dt\leq \int_0^x dt=x.


    The point is not to evaluate the integral (you can't) but to prove it is small when \epsilon is small.

    The whole scheme is like: let \epsilon>0. Using the theorem, we have \int_\epsilon^\pi f_n(x)dx\to_n 0 hence for n\geq n_0, \left|\int_\epsilon^\pi f_n(x)dx\right|\leq \epsilon. Furthermore, \left|\int_0^\epsilon f_n(x)dx\right|\leq \epsilon. Thus, \left|\int_0^\pi f_n(x)dx\right|\leq 2\epsilon.


    So we showed that
    lim ∫ sin(nx)/(nx) dx ε + 0 = ε
    n->oo
    where ε is some FIXED number between 0 and pi.
    note: ∫ is from 0 to pi

    But this only gives an upper bound on our answer. How can we find the exact value of lim ∫ sin(nx)/(nx) dx?

    Thanks!
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    So we showed that
    lim ∫ sin(nx)/(nx) dx ε + 0 = ε
    n->oo
    where ε is some FIXED number between 0 and pi.
    note: ∫ is from 0 to pi

    But this only gives an upper bound on our answer. How can we find the exact value of lim ∫ sin(nx)/(nx) dx?

    Thanks!
    \varepsilon>0 was fixed but arbitrary. So, suppose the integral was non-zero and derive a contradiction.
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