 Solution

Interpret the signed inverse-power force as $m\mathrm{sign}(x)/|x{|}^p$, so noninteger $p$ is meaningful on either side of the origin. Write $\alpha =1/p\in (0,2)$. Exactly as in part (a), the one-star characteristic function is $1-I_{n,p}(t)/n$, whereThe integrand is bounded near zero and is at most $m^2{t}^2/(2x^{2p})$ at infinity. The condition $2p>1$ thus makes its integral finite. Independence and the logarithm calculation in part (a) give the limiting characteristic function $\exp (-I_p(t))$, withHere we substituted $u=m|t|/x^p$. The integral for $J_{\alpha }$ is finite because its integrand is $O(u^{1-\alpha })$ at zero and $O(u^{-1-\alpha })$ at infinity.

To determine the constant rather than leave it as an unevaluated integral, use the Gamma function representationThe integrand multiplied by $1-\cos u$ is nonnegative, so the Tonelli theorem permits exchanging the integrals. For $v>0$ the inner integral isConsequently, substituting $z=v^2$ and using the Euler beta function and Gamma reflection formula,Since $\Gamma (1+\alpha )=\alpha \Gamma (\alpha )$, this givesEvery factor in $c_p$ is positive in the stated range. The Lévy continuity theorem gives weak convergence of random variables because this characteristic function is continuous at zero. The limit is strictly stable: if $Z_1,Z_2$ are independent copies of it, the characteristic function of $a{Z}_1+b{Z}_2$ iswhich is the characteristic function of $(|a{|}^{\alpha }+|b{|}^{\alpha }{)}^{1/\alpha }{Z}_1$. This proves the asserted index, as well as identifying the symmetric stable distribution. In particular $p=2$ gives $c_2=\sqrt{\pi m/2}$, while $p=1$ gives $c_1=\pi m/2$, the centered Cauchy distribution with that scale.