= Solution
The expression is a <Test-function pairing with a Gaussian free field>, rather than the pointwise integral of an ordinary random function. For a real <test function> $\phi$, take the Dirichlet solution
$$
f_\phi=-2\pi\Delta^{-1}\phi,
\qquad f_\phi(x)=\int_DG(x,y)\phi(y)\,dy,
$$
using the Green-kernel identity allowed in the original PDF. In particular, $-\Delta f_\phi=2\pi\phi$. <Integration by parts> gives, initially for smooth compactly supported $u$ and then by <continuity> in the <Dirichlet energy space>,
$$
(u,f_\phi)_\nabla=\frac1{2\pi}\int_D\nabla u\cdot\nabla f_\phi
=\int_Du(x)\phi(x)\,dx.
$$
The integral functional is continuous in the energy norm: pull back to the <unit disc>, where the transformed test function has compact support, and use the <Poincaré inequality>. Hence $f_\phi$ is its energy-space representer. Define
$$
\boxed{(h,\phi):=(h,f_\phi)_\nabla.}
$$
By the <isonormal Gaussian process> definition, it has mean zero and <variance>
$$
\begin{aligned}
\mathbb E[(h,\phi)^2]
&=\|f_\phi\|_\nabla^2
=\int_D f_\phi(x)\phi(x)\,dx\\
&=\iint_{D\times D}\phi(x)G(x,y)\phi(y)\,dx\,dy.
\end{aligned}
$$
The logarithmic singularity is locally integrable, so this <variance> is finite for smooth compactly supported test functions. Thus
$$
\boxed{(h,\phi)\sim N\!\left(0,\iint\phi(x)G(x,y)\phi(y)\,dx\,dy\right).}
$$
The local TeX corrupted the double integral and the bracketed Green-kernel identity; the PDF supplies the identity above and does not assert an infinite <variance>.
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