According to Newton's law of universal gravitation, a material point $P$ with mass $m$ attracts a material point $P_0$ with mass $m_0$ with a force directed from $P_0$ towards $P$, of size $k\cfrac{mm_0}{r^2}$ (further for simplicity k=1).
Note that when the point $P_0$ (located at the point $(0, 0, 0)$) of mass $m_0$, is attracted by the point $P_i$ (located in the point $(x_i, y_i, z_i)$) of mass $m_i$, is the force acting between them we express as $$\dfrac{m_om_i}{r_i^3}(x_i,y_i,z_i),$$ where $r_i=\sqrt{(x_i^2+y_i^2+z_i^2)}$.
In the situation where the point $P_0$ is attracted by points $P_1, P_2, \ldots, P_n$ with masses $m_1, m_2,\ldots,m_n$ and the attracting mass has a continuous distribution on the curve $\sigma:[a,b]\rightarrow\mathbb{R}^3, \sigma(t)=(\sigma_1(t),\sigma_2(t),\sigma_3(t))$, the curve is divided into small segments by points $a=t_0<t_1<\ldots<t_n=b$. Then we write the force of attraction as the sum of the vectors
$$F=\sum_i \dfrac{m_0m_i}{r_i(\sigma(t_i))}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i)).$$
Also $$m_i=\int_{t_{i-1}}^{t_i}f(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i))\sqrt{\sigma_1'(t_i)^2+\sigma_2'(t_i)^2+\sigma_3'(t_i)^2}dt.$$
Then $$F=\sum_i \dfrac{m_0\cdot \int_{t_{i-1}}^{t_i}f(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i))\sqrt{\sigma_1'(t_i)^2+\sigma_2'(t_i)^2+\sigma_3'(t_i)^2}dt}{r_i(\sigma(t_i))}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i)).$$
How to show that this sum goes to a vector value integral? How to find this integral?