Proving the Riemann-Lebesgue Lemma for Cosine and Infinite Integrals

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

The discussion focuses on proving the Riemann-Lebesgue Lemma for cosine integrals, specifically whether the result that the integral of a piecewise continuous function multiplied by sine approaches zero as ω approaches infinity also holds for cosine. Participants confirm that cosine, being a horizontal shift of sine, implies that the lemma should apply. The conversation highlights the importance of continuity and suggests that the proof can leverage properties of integrals involving continuous functions, particularly through the use of step functions or cruder estimates.

PREREQUISITES
  • Understanding of the Riemann-Lebesgue Lemma
  • Knowledge of piecewise continuous functions
  • Familiarity with integrals involving trigonometric functions
  • Basic concepts of continuity and discontinuities in calculus
NEXT STEPS
  • Study the proof of the Riemann-Lebesgue Lemma in detail
  • Explore properties of integrals involving step functions
  • Learn about the implications of continuity in piecewise functions
  • Investigate the behavior of integrals with oscillatory functions as parameters approach infinity
USEFUL FOR

Mathematics students, particularly those studying real analysis or advanced calculus, as well as educators and researchers interested in the applications of the Riemann-Lebesgue Lemma in Fourier analysis.

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
3K