Let$$K_{a,b}(x)=\int_{0}^{\infty}{\psi({\xi})\xi^{-b}e^{ix\xi\pm \xi^{a}}d\xi}\quad a,b>0$$ where $\psi\in C^{\infty}$, equals to $0$ when $\xi<\frac{1}{2}$, and equals to $1$ when $\xi>1$.
oscillatory integrals in one dimension
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real-analysis
harmonic-analysis
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0With this assumption on $\psi$, it's not a continuous function. – 2012-09-07
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0Why don't you get rid of the $\psi$ function and just make the integral from 1 to $\infty$? – 2012-09-07
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0Can you confirm that you're trying to evaluate $$ K_{a,b}\left(x\right)=\int_{1}^{\infty} dy \ y^{-b} \exp\left(i x y\right) \exp\left(\pm y^a\right) $$ for $a,b > 0$? – 2012-09-07
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0@Davide Giraudo:sorry,I have fixed it – 2012-09-08
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0@ Eric Angle:right,it's essentially what you have written down. – 2012-09-08