The purpose of this post is to discuss the Riemann-Lebesgue Lemma, why it makes sense, and some interesting applications among which we have , the proof of which, in my opinion, explains a lot more clearly why it should equal .

Recently I have grown quite fond of the Riemann-Lebesgue Lemma, the Fourier Transform and Fourier Series, through studying Tauberian Theorems in Analytic Number Theory, and Partial Differential Equations.

The Riemann-Lebesgue lemma states that

for quite a general set of functions .

In the case of continuous functions the interpretation is that has periods of length proportional to , in which, being continuous, the variation in is really small, while the integral of in this periods is . Thus we will be adding roughly for each cycle, being the change in during the cycle. Adding these up leaves us with something of the order of , which can be made arbitrarily small as .

Lemma 1Let be continuous, then we have

*Proof:* Observe that is uniformly continuous on the compact set . Thus, given an arbitrary there is such that whenever .

Next consider so that , we partition the integration intervals into pieces of size less than

where the are a short-hand for .

Here it is clear that the first and the last integrals tend to as because

which , and similarly for the other one. Thus, for a certain with we may assume that implies that these two integrals have absolute value less than .

As for the case of with integer, note that (it goes through a whole cycle!) implies

where the last inequality follows from the fact that the distance to the left-most point of the interval is less than and so .

Hence, collecting terms, we get

for .

**Example 1.** We will prove that (assuming its existence, which may be shown by the alternating series test):

Let us denote this integral by . Consider then

which clearly satisfy and . Setting we get

Note that almost fits into the Riemann-Lebesgue, but is not continuous on , it tends to infinity close to . To remedy this problem (somewhat surprisingly) we look first at

here we run into trouble again, but at the points and instead of .

Why insist with ? Well, in fact, it is well-known that

for a proof look at below.

To finally eliminate the discontinuities, we note that

can be extended continuously (by considering the limits at ) to .

The choice should not be surprising as

when . In fact we are eliminating the contribution from the simple poles!

We are very lucky because

while

Thus

and considering odd we have

which shows by the Riemann-Lebesgue lemma.

The Riemann-Lebesgue lemma is more general actually.

Lemma 2Consider where is the -algebra of the Lebesgue-measurable sets and is the Lebesgue measure. Then

where is to be interpreted as , the Lebesgue integral.

*Proof:* The continuous functions are dense in under the Lebesgue measure .Thus it follows that, given a measurable function we have a sequence of continuous functions such that so that, decomposing as we have

from where the result follows upon making and sufficiently large in order.

Of course, the proofs read verbatim when we change for any bounded periodic-function , since the principle of the proof stays the same.

Lemma 3Consider where is the -algebra of the Lebesgue-measurable sets and is the Lebesgue measure. Let be a bounded periodic measurable function on , with period , such that . Then

where is to be interpreted as , the Lebesgue integral.

Another possible extension is the case in which we have a continuously differentiable function , in such a case we may say much more about the rate of decay.

Lemma 4Consider . Then

as

*Proof:* Applying integration by parts

and therefore

thus proving the result.

**Example 2.** We will prove that

for , and that this convergence is uniform on for any .

- First show that .
- Integrate it in from to to get
- To make the integrand of the form with differentiable when , we must remove the pole of at . Consider
then is holomorphic on because we removed the pole, in particular it is continuously differentiable on for any .

- Thus
is when .

- Observe that, by what we know from Example 1, we have
and here when . Indeed

thus, integrating by parts, we get .

**Proof of ****.** Somewhat surprisingly, the computation of

boils down to a matter of evaluating a particular coefficient of a polynomial!

Let us write , so that we get

Here observe that so that

Note that we have

and therefore, by factoring from each of the factors, we get

where the notation denotes the coefficient of in .

For instance, for , we observe

as the term appears in the sum if and only if is even.

**Remark. **The integral may also be computed by induction and the use of trigonometric identities. However, I feel that the proof above is much more illuminating.

**Exercise. **Use the result from Example 2. to prove that .