= Solution
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:
$$
q_n(t)=\mathbb E\exp\!\left(\frac{itm\operatorname{sign}(X_1)}{X_1^2}\right)
=\frac1n\int_0^n\cos\!\left(\frac{mt}{x^2}\right)\,dx
=1-\frac{I_n(t)}n,
\qquad I_n(t)=\int_0^n\left(1-\cos\!\left(\frac{mt}{x^2}\right)\right)\,dx.
$$
Near zero the integrand is bounded by $2$. For large $x$ it is at most $m^2t^2/(2x^4)$. The <monotone convergence theorem> therefore gives the finite <integral>
$$
I_n(t)\longrightarrow I(t)=\int_0^\infty\left(1-\cos\!\left(\frac{mt}{x^2}\right)\right)\,dx.
$$
<Independence> 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> gives
$$
n\log\left(1-\frac{I_n(t)}n\right)=-I_n(t)+O(n^{-1})\longrightarrow-I(t).
$$
The <logarithm> is legitimate for all sufficiently large $n$.
For $t\ne0$, substitute $u=m|t|/x^2$. The <cosine integral for a symmetric stable exponent>, evaluated directly in part (b), gives
$$
I(t)=\frac{\sqrt{m|t|}}2\int_0^\infty\frac{1-\cos u}{u^{3/2}}\,du
=\sqrt{\frac{\pi m}{2}}\,|t|^{1/2}.
$$
At $t=0$ 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 $1/2$ with
$$
\boxed{\phi(t)=\exp\!\left(-\sqrt{\frac{\pi m}{2}}\,|t|^{1/2}\right).}
$$
The 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>.
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