= Solution
Interpret the signed inverse-power force as $m\operatorname{sign}(x)/|x|^p$, 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 $1-I_{n,p}(t)/n$, where
$$
I_{n,p}(t)=\int_0^n\left(1-\cos\!\left(\frac{mt}{x^p}\right)\right)\,dx.
$$
The integrand is bounded near zero and is at most $m^2t^2/(2x^{2p})$ at infinity. The condition $2p>1$ thus makes its <integral> finite. <Independence> and the <logarithm> calculation in part (a) give the limiting <characteristic function> $\exp(-I_p(t))$, with
$$
I_p(t)=\alpha m^\alpha|t|^\alpha J_\alpha,
\qquad J_\alpha=\int_0^\infty(1-\cos u)u^{-1-\alpha}\,du.
$$
Here 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> representation
$$
u^{-1-\alpha}=\frac1{\Gamma(1+\alpha)}\int_0^\infty v^\alpha e^{-vu}\,dv.
$$
The integrand multiplied by $1-\cos u$ is nonnegative, so the <Tonelli theorem> permits exchanging the <integrals>. For $v>0$ the inner <integral> is
$$
\int_0^\infty e^{-vu}(1-\cos u)\,du
=\frac1v-\frac{v}{1+v^2}=\frac1{v(1+v^2)}.
$$
Consequently, substituting $z=v^2$ and using the <Euler beta function> and <Gamma reflection formula>,
$$
\begin{aligned}
J_\alpha
&=\frac1{\Gamma(1+\alpha)}\int_0^\infty\frac{v^{\alpha-1}}{1+v^2}\,dv\\
&=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(1+\alpha)}
=\frac{\pi}{2\Gamma(1+\alpha)\sin(\pi\alpha/2)}.
\end{aligned}
$$
Since $\Gamma(1+\alpha)=\alpha\Gamma(\alpha)$, this gives
$$
\boxed{c_p=\frac{\pi m^{1/p}}{2\Gamma(1/p)\sin(\pi/(2p))},\qquad
\phi(t)=e^{-c_p|t|^{1/p}}.}
$$
Every 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 $aZ_1+bZ_2$ is
$$
\exp\{-c_p(|a|^\alpha+|b|^\alpha)|t|^\alpha\},
$$
which 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.
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