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I was asked to prove that

$$\lim_{n\to\infty} \int_{0}^{1} \exp(i\cdot n\cdot p(x))\;dx =0 $$

for nonconstant real polynomial $p(x)$.

if $p(x)$ is of degree $1$... It reduces to Riemann-Lebesgue lemma.

I think similar motivation will work... but not able to

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You may want to proofread your question whether the limit symbol is omitted or not. –  sos440 Oct 2 '12 at 16:35
    
Did you meant to have a limit in there? –  copper.hat Oct 2 '12 at 16:35
    
oh.. I made a huge mistake. –  Detectives Oct 2 '12 at 17:15
    
we need a limit n goes to infinity. –  Detectives Oct 2 '12 at 17:15
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1 Answer

up vote 0 down vote accepted

Suppose first that $p'$ is nonzero on $(0,1)$, so that $p$ has an inverse $q$ on $[0,1]$ which is differentiable everywhere except possibly at the endpoints. Then making the substitution $u=p(x)$ yields $$\int_0^1 e^{i \, n \, p(x)} \, dx=\int_{p(0)}^{p(1)} e^{i \, n \, u} \, q'(u) \, du \, .$$ Since $q'$ is the derivative of a bounded function, it is integrable. So the ordinary Riemann-Lebesgue lemma applies.

In general, $p'$ may not be nonzero on the entire interval. But since $p$ is a polynomial, we can split $[0,1]$ up into some finite collection of intervals over which $p'$ is nonzero, and make the above substitution separately on each of them...

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I never thought change of variable will work.. Thank you! –  Detectives Oct 3 '12 at 16:03
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