Proving the Riemann-Lebesgue Lemma for Cosine and Infinite Integrals

  • Thread starter kingwinner
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In summary: This is called integration by parts. Can you do that? In summary, the Riemann-Lebesgue Lemma states that if g is piecewise continuous on the interval [a,b], then the integral of g(t) sin(ωt) from a to b approaches 0 as ω approaches infinity. This can also be applied to the integral of g(t) cos(ωt) from a to b, by shifting the integrand horizontally. To prove this, we can use integration by parts and the assumption that g is continuous.
  • #1
kingwinner
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Homework Statement



Riemann-Lebesgue Lemma:
If g is piecewise continuous on the interval [a,b], then
b
∫ g(t) sin(ωt) dt -> 0 as ω->∞
a

[this is quoted directly from my textbook]

(i) Now assuming this result, is it possible to prove from this result that
b
∫ g(t) cos(ωt) dt -> 0 as ω->∞ ??
a
I think it also works for cosine becuase cosine is just a horizontal shift of sine, but does this imply that the second result(with cos) is an immediate consequence of the first(with sin)? How can we prove this?


(ii) Also, if we have


-∞
, is the lemma above still true?

Homework Equations


Riemann-Lebesgue Lemma

The Attempt at a Solution


N/A

Any help is appreciated! :)
 
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  • #2
kingwinner said:
I think it also works for cosine becuase cosine is just a horizontal shift of sine, but does this imply that the second result(with cos) is an immediate consequence of the first(with sin)? How can we prove this?
Well, can we prove it by shifting horizontally?

Sure, there are some issues -- but maybe we can prove those issues converge to zero?

(WLOG, we can assume g is actually continuous)
 
  • #3
(i)

b
∫ g(t) sin(ωt) dt -> 0 as ω->∞
a

implies

b
∫ g(t) cos(ωt - pi/2ω) dt -> 0 as ω->∞
a

But from here, how can we prove that
b
∫ g(t) cos(ωt) dt -> 0 as ω->∞ ??
a
 
  • #4
Is there absolutely nothing you can think to do to that integral or that integrand?
 
  • #5
There is a really short solution to this via step functions, but if you don't have that available, then you can use cruder estimates. As Hurkyl mentioned, we may assume g is continuous. Understanding why this doesn't change the problem is the key to figuring out a good first step. Remember sectionally continuous just means continuous except for a finite number of jump discontinuities, and we can certainly take the definite integral over a continuous interval with discontinuous endpoints, since the endpoints don't really matter.
 
  • #6
You have an integrand that is the product of two functions. There is a way to rewrite that integral, by splitting up the parts.
 

What is the Riemann-Lebesgue Lemma?

The Riemann-Lebesgue Lemma is a mathematical theorem that describes the behavior of the Fourier transform of a function as the frequency approaches infinity. It essentially states that the Fourier transform of a function approaches zero as the frequency increases, and this result is used in many areas of mathematics and science.

Who discovered the Riemann-Lebesgue Lemma?

The Riemann-Lebesgue Lemma was first proved by the German mathematician Bernhard Riemann in the mid-19th century. However, it was later refined and popularized by the French mathematician Henri Lebesgue in the early 20th century, hence the name "Riemann-Lebesgue" Lemma.

What is the significance of the Riemann-Lebesgue Lemma in mathematics?

The Riemann-Lebesgue Lemma has many applications in mathematics, particularly in analysis and harmonic analysis. It is used to prove important theorems such as the Central Limit Theorem and the Dirichlet's Theorem on Diophantine approximations. It also plays a crucial role in the study of Fourier series and Fourier transforms.

What is the intuition behind the Riemann-Lebesgue Lemma?

The Riemann-Lebesgue Lemma can be intuitively understood as follows: as the frequency of a function increases, the contribution of the function to its Fourier transform decreases. In other words, high-frequency components of a function "smooth out" when transformed into the frequency domain, and this is what the lemma formally states.

Are there any generalizations of the Riemann-Lebesgue Lemma?

Yes, there are several generalizations of the Riemann-Lebesgue Lemma. One of the most well-known is the Riemann-Lebesgue Theorem, which extends the lemma to Lebesgue integrable functions on the real line. There are also generalizations to higher dimensions, as well as versions for different types of Fourier transforms, such as the Laplace transform and the Mellin transform.

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