# Riemann-Lebesgue Lemma

• Nov 10th 2009, 07:13 PM
kingwinner
Riemann-Lebesgue Lemma
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?

Any help is appreciated! :)

[note: also under discussion in s.o.s. math cyberboard]
• Nov 11th 2009, 12:05 PM
Opalg
Quote:

Originally Posted by kingwinner
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?

I think that method basically works, but the details get a bit messy. In fact, if you make the substitution $u=t - \tfrac\pi{2\omega}$ in the integral then you get

$\int_a^b g(t)\sin\omega t\,dt = \int_{a-(\pi/2\omega)}^{b-(\pi/2\omega)} g\bigl(u+\tfrac\pi{2\omega}\bigr) \sin\bigl(\omega u+\tfrac\pi2\bigr)\,du = \int_{a-(\pi/2\omega)}^{b-(\pi/2\omega)} g\bigl(u+\tfrac\pi{2\omega}\bigr) \cos\omega u\,du$.

Now suppose that g(t) is continuous for t in [a,b]. Then it is uniformly continuous, so that last integral will be close to what it would be with g(u) in place of $g\bigl(u+\tfrac\pi{2\omega}\bigr)$ (provided that $\omega$ is large). Also, the difference caused by shifting the limits of integration back to a and b will not affect the integral much. Thus if $\int_a^bg(t)\sin\omega t\,dt$ is small then so is $\int_a^bg(t)\cos\omega t\,dt$.

Converting that vague talk about things being small into a precise argument with epsilons will be messy but not inherently difficult. (In practice, however, I think it would be easier to repeat the proof of the sine result, replacing sines by cosines, rather than trying to deduce one result from the other.)

If you only know that g is piecewise continuous (that is, continuous except for a finite number of jump discontinuities) then you can split the interval [a,b] into a finite number of subintervals on which g is continuous, and apply the above argument to each of them.

Quote:

Originally Posted by kingwinner
(ii) Also, if we have

-∞
, is the lemma above still true?

Yes, provided that $\textstyle\int_{-\infty}^\infty|g(t)|\,dt < \infty$. That condition is necessary in order for the integrals to converge. The proof than uses the fact that g(t) must be small outside some bounded interval. Inside the interval, the previous argument applies.