Proving the Riemann-Lebesgue Lemma for Cosine and Infinite Integrals

  • Thread starter Thread starter kingwinner
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the Riemann-Lebesgue Lemma, specifically its implications for integrals involving cosine functions. The original poster questions whether the lemma's result for sine functions can be extended to cosine functions and explores the validity of the lemma in the context of infinite integrals.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between sine and cosine functions, considering whether the result for sine can be directly applied to cosine. Questions arise about the implications of horizontal shifts and the convergence of certain integrals.

Discussion Status

Participants are actively exploring the connections between the sine and cosine integrals, with some suggesting that continuity assumptions may simplify the problem. There is a recognition of potential issues that need to be addressed, but no consensus has been reached on a definitive approach.

Contextual Notes

Some participants note that the lemma's applicability may depend on the continuity of the function g, and there are discussions about the implications of using step functions or cruder estimates. The original poster also raises questions about the validity of the lemma in the context of infinite integrals.

kingwinner
Messages
1,266
Reaction score
0

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 because 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! :)
 
Physics news on Phys.org
kingwinner said:
I think it also works for cosine because 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)
 
(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
 
Is there absolutely nothing you can think to do to that integral or that integrand?
 
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.
 
You have an integrand that is the product of two functions. There is a way to rewrite that integral, by splitting up the parts.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
9
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K