$$\delsq \psi = \sin\theta {\cal L}[g(r)]\  \ {\rm where}\ \ {\cal L} [g(r)] \equiv  \frac{1}{r}\frac{\partial\ }{\partial r}\left(r\frac{\partial g}{\partial r} \right)- 
\frac{g}{r^2}$$ and the pressure equation becomes: $$\del p = \mu U \left\{-\ervec \frac{1}{r} \cos\theta {\cal L}[g(r)] + \ethvec \sin\theta \frac{\partial\ }{\partial r}{\cal L}[g(r)]\right\}$$ We do the integration from infinity along $\theta = \pi/2$ and then work around the circle $r=1$. Carrying this out we obtain $$p(1,\theta) = p_{\infty} - \mu U \cos \theta \frac{\partial\ }{\partial r} {\cal L}[g(r)]|_{r=1}$$ We now evaluate the total integral for the force per unit length to obtain $$D_0 = \mu U \pi \left[\frac{\partial\ }{\partial r}{\cal L}[g(r)]-\frac{\partial^2 \!g(r)}{\partial r^2} \right]$$ The drag coefficient for such a 2D problem is defined following Tritton (1959) as $C_D = |D_0/(\rho U^2)|$ and is now easily seen to be given by $$C_D = \frac{\pi}{R_e} \left|\frac{\partial\ }{\partial r}{\cal L}[g(r)]-
\frac{\partial^2\!g(r)}{\partial r^2} \right|$$ evaluated on $r=1$. Given the vanishing of $g(1)$ and $g'(1)$ this simplifies further to $$C_D = \frac{\pi}{R_e} |g^{(3)}(1) |$$ In the case given by the MAE approach the stream function is given in the neighbourhood of the cylinder by Eq. (1) and then $g$ and $C_D$ are given by $$g(r) = C\left(r \log r -\frac{r}{2} + \frac{1}{2r}\right),\ \ C_D 