# Homework Help: Riemann-Lebesgue Lemma

1. Nov 10, 2009

### kingwinner

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
Riemann-Lebesgue Lemma

3. The attempt at a solution
N/A

Any help is appreciated! :)

2. Nov 10, 2009

### Hurkyl

Staff Emeritus
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. Nov 10, 2009

### kingwinner

(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. Nov 10, 2009

### Hurkyl

Staff Emeritus
Is there absolutely nothing you can think to do to that integral or that integrand?

5. Nov 10, 2009

### snipez90

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. Nov 11, 2009

### Quantumpencil

You have an integrand that is the product of two functions. There is a way to rewrite that integral, by splitting up the parts.