How to Prove the Limit of an Integral Involving Sine and a Function?

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SUMMARY

The discussion focuses on proving the limit of an integral involving the sine function, specifically the expression \lim_{N\to\infty} \int\limits_0^{NA} \frac{\sin(x)}{2N\sin(\frac{x}{2N})} dx = \frac{\pi}{2} for 0. The user successfully applies complex analysis and series to establish foundational results, including \int\limits_0^{\infty} \frac{\sin(x)}{x}dx = \frac{\pi}{2}. The transformation of the integral into a limit of integrals over a fixed interval is achieved using the Dirichlet Kernel, and the Riemann-Lebesgue Lemma is identified as a key tool for solving the problem. The user concludes with a newfound understanding of Fourier analysis techniques.

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of the Dirichlet Kernel
  • Familiarity with the Riemann-Lebesgue Lemma
  • Basic techniques of Fourier analysis
NEXT STEPS
  • Study the properties and applications of the Dirichlet Kernel
  • Learn about the Riemann-Lebesgue Lemma in detail
  • Explore advanced techniques in Fourier analysis
  • Investigate the second mean value theorem and its applications in integrals
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Mathematicians, students of advanced calculus, and anyone interested in integral calculus and Fourier analysis techniques.

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I know how to prove

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

using complex analysis, and I know how to prove

[tex] \lim_{N\to\infty} 2N\sin\Big(\frac{x}{2N}\Big) = x[/tex]

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

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

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

[tex] <br /> 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<br /> [/tex]

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:

http://en.wikipedia.org/wiki/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)):

http://www.math.washington.edu/~morrow/335_07/rl_lemma.pdf

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.
 
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

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

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:
 
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