= Solution
The <product rule>, the <Laplace equation> and $\mathbf u=\nabla\phi$ give
$$
\boxed{\nabla\cdot(\phi\nabla\phi)=|\nabla\phi|^2+\phi\nabla^2\phi=|\mathbf u|^2.}
$$
First truncate the exterior fluid domain by a large <circle> $C_R$. Let $\mathbf n_B$ point outward from the body into the fluid. The outward normal of the exterior fluid domain at the body is $-\mathbf n_B$. The planar <divergence theorem>, equivalently <Green's theorem>, therefore gives
$$
\int_{D_R}|\mathbf u|^2\,dA
=\oint_{C_R}\phi\phi_r\,ds-\oint_{\partial B}\phi\,\partial_{n_B}\phi\,ds.
$$
Orient $\partial B$ counterclockwise around the body. Then $\mathbf n_B\,ds=(dy,-dx)$ and
$$
\partial_{n_B}\phi\,ds=\phi_x\,dy-\phi_y\,dx
=\psi_y\,dy+\psi_x\,dx=d\psi.
$$
For the finite-energy exterior rigid-body flow intended here, the outer boundary term vanishes as $R\to\infty$. In particular, zero net <volume flux> and zero <circulation> eliminate the $1/z$ term in the exterior <Laurent series> of the <complex velocity>. Thus $\mathbf u=O(R^{-2})$ and, after choosing the additive potential constant, $\phi=O(R^{-1})$. The outer integral is then $O(R^{-2})$.
Multiplying by half the constant <mass density> gives the <boundary integral for kinetic energy of planar potential flow>:
$$
\boxed{T=\frac\rho2\int_{\mathbb R^2\setminus B}|\mathbf u|^2\,dA
=-\frac\rho2\oint_{\partial B}\phi\,d\psi.}
$$
The minus sign comes from the fluid's inner-boundary orientation, rather than a negative physical <kinetic energy>.
There is a necessary qualification to the printed hypotheses. Decaying <velocity> and zero <circulation> alone do not ensure a finite <kinetic energy>: $\phi=A\log r$ gives $\mathbf u=A\mathbf e_r/r$, zero <circulation>, and logarithmically divergent exterior <kinetic energy>. It carries a nonzero <volume flux> from the body and is not the flow of an impermeable rigid body. The <finite-energy condition for planar exterior potential flow> makes explicit the source-free assumption needed to discard the outer boundary term.
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