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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 $\xi>1$.

As far as I know, when $0<a<1$, $a=1$,and $a>1$,the behaviors of $K_{a,b}$ are quite different (for proper choice of $b$, the singularity lies in $x=0$,$|x|=1$, $x=\infty$ respectively). I want to know the reason behind this and I also want to know how to evaluate such kind of integrals.

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With this assumption on $\psi$, it's not a continuous function. –  Davide Giraudo Sep 7 '12 at 21:43
    
Why don't you get rid of the $\psi$ function and just make the integral from 1 to $\infty$? –  Eric Angle Sep 7 '12 at 21:59
    
Can 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 $a,b > 0$? –  Eric Angle Sep 7 '12 at 22:06
    
@Davide Giraudo:sorry,I have fixed it –  sun Sep 8 '12 at 0:01
    
@ Eric Angle:right,it's essentially what you have written down. –  sun Sep 8 '12 at 0:06

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