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Integrand approaching sin(x)/x

  1. Aug 9, 2009 #1
    I know how to prove

    \int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}

    using complex analysis, and I know how to prove

    \lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x

    using series. I have some reason to believe, that if [itex]0<A<\pi[/itex], then

    \lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2},

    but don't know how to prove this. Anyone knowing how to accomplish this?
  2. jcsd
  3. Aug 9, 2009 #2
    Let I denote the integral you are evaluating. Substitute t = x/N (N*dt = dx) to get


    I = \frac{1}{2} \lim_{N\to\infty} \int\limits_0^{A} \frac{\sin(Nt)}{\sin(\frac{t}{2})} \, dt = \frac{1}{2} \lim_{n\to\infty} \int\limits_0^{A} \frac{\sin((n+\frac{1}{2})t)}{\sin(\frac{t}{2})} \, dt


    We transformed the original expression I into a limit of integrals over a fixed interval. The integrand of the last expression is the Dirichlet Kernel:


    Using the identity from the wikipedia article, find out what happens if we let [tex]A = \pi[/tex]

    A clever application of the Riemann-Lebesgue Lemma (Theorem 1 (2)):


    solves the problem. It is not obvious what f(t) should be, but remember you have the dirichlet kernel and the improper sine integral you guessed would equal I.

    This outlined approach amounts to nothing more than an elementary evaluation of the improper sinx / x integral.
  4. Aug 13, 2009 #3
    Actually I was originally trying to read stuff about Dirichlet's kernel (and I encountered the Riemann-Lebesgue lemma at the same time), but I didn't feel like understanding the proofs, then tried to do them my own way, and arrived at the problem that I described in the first post. Your advise to apply Dirichlet's kernel results merely directed me back to my original problems. Anyway, you were right that I should return back towards the books, since my problem in the first post was not in the right direction.

    It seems that I have now finally understood some basic techniques of the Fourier analysis. The big problem (to which my previous problem can be reduced to), is the question

    \lim_{A\to\infty} \int\limits_0^{R} \frac{\sin(Ax)}{x} f(x)dx = \frac{\pi}{2}f(0)?

    I have now understood how to prove this if [itex]f:[0,R]\to\mathbb{R}[/itex] is monotonic and differentiable, using the second mean value theorem (or some version of it) at the heart of the proof. A great achievement! Finally, years of confusion have reached their end. :cool:
    Last edited: Aug 13, 2009
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