= Solution
Use the <Newtonian fluid stress tensor> $\sigma=-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The <Stokes equations> for a point <force> on the fluid are $\nabla\cdot u=0$ and $\nabla\cdot\sigma+F\delta(x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the <Unscaled Papkovich–Neuber representation> gives the rotationally covariant decaying solution. Specifically, take $\Phi=-F/(8\pi\mu r)$ and $\chi=0$. Differentiation yields
$$
\boxed{u_i=J_{ij}F_j,\quad J_{ij}(x)=\frac1{8\pi\mu}\left(\frac{\delta_{ij}}r+\frac{x_ix_j}{r^3}\right),\qquad p=\frac{F\cdot x}{4\pi r^3}.}
$$
This satisfies the homogeneous <Stokes equations> for $r>0$. Its <stress> is
$$
\boxed{\sigma_{ij}=K_{kij}F_k,\qquad K_{kij}(x)=-\frac{3x_kx_ix_j}{4\pi r^5}.}
$$
To verify the point-force normalization, on a sphere $r=\varepsilon$ the <traction> is $\sigma n=-3(F\cdot n)n/(4\pi\varepsilon^2)$. Since $\int_{S^2}n_in_j\,d\Omega=4\pi\delta_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied <force> $F$, fixing the normalization of the <Stokeslet>. Its radial <velocity> is $(F\cdot n)/(4\pi\mu r)$, so
$$
\boxed{\int_{|x|=r}u\cdot n\,dS=\frac r{4\pi\mu}F\cdot\int_{S^2}n\,d\Omega=0.}
$$
There is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.
For two Stokes flows with the same <viscosity> and body <forces> $f^{(1)},f^{(2)}$, the <Lorentz reciprocal theorem> including body <forces> is
$$
\boxed{\int_{\partial V}\big(u^{(1)}\cdot\sigma^{(2)}n-u^{(2)}\cdot\sigma^{(1)}n\big)dS
=\int_V\big(u^{(2)}\cdot f^{(1)}-u^{(1)}\cdot f^{(2)}\big)dV.}
$$
Indeed, apply the <divergence theorem> to $u_i^{(1)}\sigma_{ij}^{(2)}-u_i^{(2)}\sigma_{ij}^{(1)}$. The cross-gradient terms cancel: <incompressibility> removes each <pressure> contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two <stress> divergences give the body-force terms shown.
Choose flow 1 to be the desired body-force-free field and flow 2 to be a <Stokeslet> with arbitrary <force> vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ gives
$$
\boxed{u_i(y)=\int_{\partial V}J_{ij}(y-x)\sigma_{jk}(x)n_k\,dS
+\int_{\partial V}u_j(x)K_{ijk}(y-x)n_k\,dS.}
$$
For $y$ outside the volume there is no point <force> in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi/(4\pi)$. The result is the <boundary integral representation of Stokes flow> with right side $\tfrac12u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary <principal value>. A nonsmooth corner would have a different solid-angle coefficient.
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