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Consider the standard Lévy-Stable (or Alpha Stable) distribution $S(\alpha,\beta, \mu, \sigma)$ where $\alpha$ is the tail exponent, $1 \leq \alpha \leq 2 $. Picking the symmetric case with $0$ mean (that is, $\beta =0$, $\mu =0$ the characteristic function becomes $C(t)= e^{-\left| t \sigma \right| ^{\alpha }}$.

We can numerically extract the mean absolute deviation by inverting the characteristic function. From the shape of the distribution shown below in the graph, and, where $X$ is the r.v., from the property that $E(|X|)_{\alpha=2}= \frac{2 }{\sqrt{\pi }} \sigma$ (the only explicit solution we get) and $E(|X|)_{\alpha=1}= \infty$ ( the Cauchy case), a good guess is: $$E(|X|)= \left(\frac{a}{\log(\alpha)}\right)^{f(\alpha)} \sigma,$$ where $a=\frac{2 \log (2)}{\sqrt{\pi }}$ and $f ( \alpha)= 1$ when $\alpha =2$:

I wonder if someone knows of an explicit expression.

The mumerically derived mean absolute deviation as a function of tail exponent.

Added Comment: By tinkering I figured out we could track the function with $E(|X|)=\frac{a_1 \sqrt{\alpha }}{(a_2+\alpha)\, B\left(a_3 \alpha, a_4\right)}$,where $B$ is the Beta function, but the number of parameters is high.

Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$. Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$, $$ E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s $$

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Note: Copied from question. Delete I guess if NNT answers his own question.

Answer: Thanks to help it looks like $E(|X|)= \frac{2 \Gamma \left(1-\frac{1}{\alpha }\right)}{\pi } \sigma$. Zolotarev: where $m(X)$ is the median and $\lambda=\sigma^\alpha$, $$ E|X- m(X)|^s= \frac{1}{\pi} 2^{1-s/2} (2 \lambda)^{s/\alpha} \Gamma(1-\frac{s}{\alpha}) \Gamma(s) \sin\frac{\pi}{2} s $$

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