= Solution
<Irrotational flow> means $\nabla\times\boldsymbol u=0$; <incompressible flow> means $\nabla\cdot\boldsymbol u=0$. The given <streamfunction> automatically satisfies the second condition, since $\partial_x\psi_y-\partial_y\psi_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\boldsymbol n=(dy/ds,-dx/ds)$. Then
$$
\boldsymbol u\cdot\boldsymbol n\,ds=\psi_y\,dy+\psi_x\,dx=d\psi,
\qquad
\boxed{\int_{\boldsymbol x_1}^{\boldsymbol x_2}\boldsymbol u\cdot\boldsymbol n\,ds
=\psi(\boldsymbol x_2)-\psi(\boldsymbol x_1)}.
$$
The opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The <volume flux> between two <streamlines> is thus their <streamfunction> difference. A closed <contour> has zero net flux, expressing <mass conservation> for constant-density flow without a source inside.
For the channel, a rigid boundary has zero normal velocity and hence constant <streamfunction>. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi=-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta\psi$, so irrotationality gives $\Delta\psi=0$.
Write $\psi=-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. <Separation of variables> gives the <harmonic sine mode in a half-strip> $e^{-n\pi x}\sin n\pi y$. The <Fourier sine series> coefficients of the left boundary are
$$
b_n=2\int_0^1[-m(1-y)]\sin(n\pi y)\,dy=-\frac{2m}{n\pi}.
$$
Therefore the <corner sink in a semi-infinite channel> is
$$
\boxed{\psi(x,y)=-my-\frac{2m}{\pi}\sum_{n=1}^\infty\frac{e^{-n\pi x}\sin(n\pi y)}n}.
$$
For $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, $\sum_{n\ge1}\sin(n\pi y)/n=\pi(1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.
\b[The plus sign printed in the final series is inconsistent with the printed <boundary conditions>.] It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi\sim-2m\theta/\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.
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