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The force from one location is defined almost surely, since the uniform distribution assigns probability zero to the origin. Symmetry of that uniform distribution makes its characteristic function real:Near zero the integrand is bounded by $2$. For large $x$ it is at most $m^2{t}^2/(2x^4)$. The monotone convergence theorem therefore gives the finite integralIndependence of the locations gives the characteristic function of the total force as $q_n(t{)}^n$. For fixed $t$, $I_n(t)$ is bounded, so Taylor expansion of the logarithm givesThe logarithm is legitimate for all sufficiently large $n$.

For $t\ne 0$, substitute $u=m|t|/x^2$. The cosine integral for a symmetric stable exponent, evaluated directly in part (b), givesAt $t=0$ both sides vanish. The limiting characteristic function is continuous at zero, so the Lévy continuity theorem proves weak convergence of random variables to the symmetric stable distribution of index $1/2$ withThe coefficient reflects the spatial density $n/(2n)=1/2$. Symmetry was used in the characteristic function, without assuming that the singular force has a finite mean.

Place equal masses at points of a homogeneous Poisson point process of intensity $\rho$ on the line, and truncate their signed inverse-power forces symmetrically at radius $R$. For force $m\mathrm{sign}(x)/|x{|}^p$, with $\alpha =1/p\in (0,2)$, the characteristic exponent tends to $-2\rho {\int }_0^{\infty }(1-\cos (mt/x^p))dx$. The cosine integral for a symmetric stable exponent evaluates it as displayed. Thus the weak limit is a symmetric stable distribution. Symmetric truncation is important when the far-field force series is not absolutely summable.