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Assume that $f:[0,1] \to R$ is a smooth function. Prove that

$$\lim_{n \to \infty} \int_0^1f(x)e^{inx^3}dx = 0.$$

Attempt at solution: I think the solution may require the interchange of limit and integral. Have tried using the power series expansion for $e^{inx^3}$ but it didn't help. Can't seem to use Fourier inversion either.

Any advice would be much appreciated. Thanks.

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Suppose $f(x) = \cos(nx^3)$. Then, the real part of your integrand is always non-zero, as it is a square. Are you sure you got the problem right? –  dexter04 Dec 5 '12 at 3:27
@dexter04 the function can't depend on $n$. –  Zarrax Dec 5 '12 at 4:37

1 Answer 1

up vote 1 down vote accepted

Change variables $y = x^3$. Then you're trying to find $$\lim_{n \rightarrow \infty} \int_0^1 {1 \over 3}y^{-{2 \over 3}}f(y^{1 \over 3})e^{iny}\,dy$$ Since $f(x)$ is smooth it is bounded. Thus ${1 \over 3}y^{-{2 \over 3}}f(y^{1 \over 3})$ is in $L^1$ since $y^{-{2 \over 3}}$ is. Hence the Riemann-Lebesgue lemma implies the limit is just zero. (If you're worried about the domain of integration being $[0,1]$ instead of $[0,2\pi]$, extend ${1 \over 3}y^{-{2 \over 3}}f(y^{1 \over 3})$ to be zero on $(1,2\pi)$.)

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Zarrax - thank you for the super-quick response. Much appreciated. –  Conan Wong Dec 5 '12 at 4:44

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