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.