What is the asymptotic behavior of the Levy distribution $$P(x)=\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos qx\,dq$$ when the independent variable $x$ approaches $0$?
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2$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Sep 13, 2023 at 14:59
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$\begingroup$ for small $x$ you have $\log P(x)\propto -1/x$. $\endgroup$– Carlo BeenakkerCommented Sep 13, 2023 at 20:02
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$\begingroup$ Hello, thank you very much for your answer, are there some references please. $\endgroup$– 吴月红Commented Sep 14, 2023 at 3:44
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$\begingroup$ see en.wikipedia.org/wiki/Lévy_distribution $\endgroup$– Carlo BeenakkerCommented Sep 14, 2023 at 6:20
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$\begingroup$ Sorry, since I didn't state the definition of levy step-by-step before, the answer you gave doesn't match my question. I modified the question, can you help answer it again. $\endgroup$– 吴月红Commented Sep 14, 2023 at 8:27
1 Answer
$$P_{\alpha,\gamma}(x)=\frac{1}{\pi}\int_{0}^\infty \exp(-\gamma q^\alpha)\cos qx\,dq$$ $$=\frac{1}{\pi}\gamma^{-1/\alpha} \Gamma \left(1+\frac{1}{\alpha}\right)-\frac{1}{2\pi\alpha}x^2 \gamma^{-3/\alpha} \Gamma \left(\frac{3}{\alpha}\right)+{\cal O}(x^4).$$
To check, I take $\alpha=1$ or $\alpha=2$, when $$P_{1,\gamma}(x)=\frac{\gamma}{\gamma^2+x^2}=\frac{1}{\gamma}-\frac{x^2}{\gamma^3}+{\cal O}(x^4),$$ $$P_{2,\gamma}(x)=\frac{\sqrt{\pi } e^{-\frac{x^2}{4 \gamma}}}{2 \sqrt{\gamma}}=\frac{\sqrt{\pi }}{2 \sqrt{\gamma}}-\frac{\sqrt{\pi } x^2}{8 \gamma^{3/2}}+{\cal O}(x^4),$$ in agreement with the expansion above.
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$\begingroup$ Thank you! Did you calculate this integral with Taylor expansion? $\endgroup$– 吴月红Commented Sep 14, 2023 at 8:50
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$\begingroup$ yes, assuming $\gamma>0$ and $\alpha>0$ so that the integral converges $\endgroup$ Commented Sep 14, 2023 at 8:51
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$\begingroup$ Thank you, because q is taking from 0 to infinite ,does not Taylor expansion work? what is more ,if \alpha is less than 0 , the result of integral? $\endgroup$– 吴月红Commented Sep 14, 2023 at 9:01
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$\begingroup$ for $\alpha\leq 0$ the integral is divergent $\endgroup$ Commented Sep 14, 2023 at 10:32
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$\begingroup$ Taylor expansion works because the integral is dominated by the small-$q$ region; as a check, I added the cases $\alpha=1$ and $\alpha=2$, which can be evaluated in closed form. $\endgroup$ Commented Sep 14, 2023 at 10:37