Calculus books tend to introduce transcendental functions (trigonometric, exponential, logarithm) early. Analysis textbooks such as *Principles of Mathematical Analysis* by Rudin tend to introduce them later, because of how long it takes to develop enough of the theory of power series.

The Riemann-Lebesgue lemma involves either trigonometric or exponential functions. But the following version works with the “late transcendentals” approach.

## Transcendental-free Riemann-Lebesgue Lemma

**TFRLL**. Suppose that and are continuously differentiable functions, and is bounded. Then as .

The familiar form of the lemma is recovered by letting or .

**Proof**. By the chain rule, is the derivative of . Integrate by parts:

By assumption, there exists a constant such that everywhere. Hence , , and . By the triangle inequality,

completing the proof.

As a non-standard example, TFRLL applies to, say, for which . The conclusion is that , that is, which seems somewhat interesting. When , the factor of can be removed by applying the result to , leading to .

## What if we tried less smoothness?

Later in Rudin’s book we encounter the Weierstrass theorem: every continuous function on is a uniform limit of polynomials. Normally, this would be used to make the Riemann-Lebesgue lemma work for any continuous function . But the general form given above, with an unspecified , presents a difficulty.

Indeed, suppose is continuous on . Given , choose a polynomial such that on . Since has continuous derivative, it follows that . It remains to show that is close to . By the triangle inequality,

which is bounded by … um. Unlike for and , we do not have a uniform bound for or for its integral. Indeed, with the integrals grow linearly with . And this behavior would be even worse with , for example.

At present I do not see a way to prove TFRLL for continuous , let alone for integrable . But I do not have a counterexample either.