= Solution
For the isolated <line vortex>, use <polar coordinates> $z=re^{i\theta}$ to obtain
$$
\phi=\frac\Gamma{2\pi}\theta,\qquad
\psi=-\frac\Gamma{2\pi}\log r,\qquad
u_r=0,\quad u_\theta=\frac\Gamma{2\pi r}.
$$
Thus for nonzero $\Gamma$ the <streamlines> are circles and their counterclockwise <circulation> is $2\pi r u_\theta=\Gamma$. If $\Gamma=0$ the fluid is stationary and the <streamline> description degenerates.
Place an opposite image <vortex> at $z=-1$. The <method of images> gives
$$
\boxed{w(z)=-\frac{i\Gamma}{2\pi}\bigl(\log(z-1)-\log(z+1)\bigr)}.
$$
On $z=iy$, the distances to the two <vortex> positions agree, so $\psi=-(\Gamma/2\pi)\log(|z-1|/|z+1|)=0$. Hence the wall is a <streamline>. Directly the wall velocity has $u=0$ and
$$
v(0,y)=-\frac\Gamma{\pi(1+y^2)}.
$$
For the stipulated steady flow, <Bernoulli's equation> in the irrotational fluid outside the <vortex> core gives
$$
p(0,y)-p_\infty=-\frac12\rho v(0,y)^2
=-\frac{\rho\Gamma^2}{2\pi^2(1+y^2)^2}.
$$
The force exerted by the fluid on this wall points in the $-x$ direction for positive <pressure>. Relative to uniform <pressure> $p_\infty$, however, its change is in the $+x$ direction. Using the supplied integral, the force change per unit span is
$$
\boxed{\Delta\mathbf F_{\rm wall}=\frac{\rho\Gamma^2}{4\pi}\,\mathbf e_x}.
$$
Thus the wall is attracted \b[towards the <vortex>], with this magnitude if the other side remains at $p_\infty$. Subtracting the uniform <pressure> is essential for an infinite wall. A freely moving <vortex> would translate parallel to the wall under its image-induced velocity; the calculation here is for the stated steady <vortex> configuration, with its position held fixed.
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