= Solution
Put $\mathbf t=\boldsymbol\theta$. The <product rule> applied to $Q_{ij}=\lambda(t_it_j-\delta_{ij}/2)$ gives
$$
(\nabla\cdot Q)_j=[(\nabla\lambda)\cdot\mathbf t]t_j+\lambda(\nabla\cdot\mathbf t)t_j+\lambda[(\mathbf t\cdot\nabla)\mathbf t]_j-\frac12\partial_j\lambda.
$$
Since $\operatorname{Tr}Q^2=\lambda^2/2$, substituting this vector into the divergence elastic term proves the requested form of the local density. For the radial amplitude and tangential director, the <polar coordinates> identities are
$$
\nabla\lambda=\lambda'(r)\mathbf e_r,\qquad (\nabla\lambda)\cdot\mathbf t=0,\qquad \nabla\cdot\mathbf t=0,\qquad (\mathbf t\cdot\nabla)\mathbf t=-\frac{\mathbf e_r}{r}.
$$
Hence the <tangential integer nematic defect core> has
$$
\boxed{\nabla\cdot Q=-\left(\frac{\lambda'}2+\frac\lambda r\right)\mathbf e_r,\qquad f=\frac a2\lambda^2+\frac b4\lambda^4+\frac K2\left(\frac{\lambda'}2+\frac\lambda r\right)^2.}
$$
The $1/r$ director-curvature term survives when the amplitude is constant. Replacing the divergence norm by a full tensor-gradient norm would not give this expression; it is the PDF's divergence convention that is being evaluated.
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