= Solution
{c}
For $0<y<\pi$, define $J(y)=\int_0^\pi u_y(x,y)\,dx$. Differentiate under the <integral> and use <Laplace's equation>:
$$
J'(y)=\int_0^\pi u_{yy}(x,y)\,dx
=-\int_0^\pi u_{xx}(x,y)\,dx
=-\bigl[u_x(\pi,y)-u_x(0,y)\bigr]=0.
$$
Thus $J$ is constant. The top <Neumann boundary condition> gives $J(\pi)=0$, so
$$
\boxed{\int_0^\pi u_y(x,y^*)\,dx=0,\qquad0<y^*<\pi.}
$$
For the included endpoint $y^*=0$, the <integral> is interpreted as an <improper integral> at the bottom corners if necessary. The <Fourier series> constructed in <c> gives, on $0<x<\pi$,
$$
u_y(x,0)=-2\log(2\sin x)
+2\sum_{m=1}^\infty\frac{\tanh(2m\pi)-1}{m}\cos(2mx).
$$
The second <series> is absolutely and uniformly convergent; the first term has only integrable <logarithmic singularities>. Its <integral> is zero since $\int_0^\pi\log(2\sin x)\,dx=0$, while every <cosine> term also has zero <integral>. This proves the endpoint as well:
$$
\boxed{\int_0^\pi u_y(x,y^*)\,dx=0,\qquad0\leq y^*<\pi.}
$$
For completeness, the logarithmic <integral> follows without any endpoint differentiation. Put $I=\int_0^{\pi/2}\log(\sin x)\,dx$ and substitute $x\mapsto\pi/2-x$ to obtain $I=\int_0^{\pi/2}\log(\cos x)\,dx$. Adding and using $\sin x\cos x=\sin(2x)/2$ gives $2I=I-(\pi/2)\log2$, hence $I=-(\pi/2)\log2$ and the required full-interval result.
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