State and prove Green’s Theorem for rectangles. Then use it to prove Cauchy’s Theorem for functions analytic in a rectangle.
Solution
Let \[ R=[a,b]\times[c,d]\subseteq\RR^2 \] and orient \(\partial R\) counterclockwise.
1. Green’s theorem for a rectangle states that if \(P,Q\) have continuousfirst partial derivatives on a neighborhood of \(R\), then\[ \boxed{ \int_{\partial R}P\,dx+Q\,dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\,dA. } \]
2. The contribution of the two horizontal sides of \(\partial R\) is\[ -\iint_R P_y\,dA. \]
The bottom side is traversed from \((a,c)\) to \((b,c)\) and the top side from \((b,d)\) to \((a,d)\). Since \(dy=0\) on both sides, their total contribution is \[ \begin{aligned} &\int_a^b P(x,c)\,dx + \int_b^a P(x,d)\,dx\\ &\qquad= \int_a^b\bigl(P(x,c)-P(x,d)\bigr)\,dx\\ &\qquad= -\int_a^b\int_c^d P_y(x,y)\,dy\,dx, \end{aligned} \] where the last equality is the fundamental theorem of calculus in the \(y\)-variable. This is \[ -\iint_RP_y\,dA. \]
3. The contribution of the two vertical sides of \(\partial R\) is\[ \iint_R Q_x\,dA. \]
The right side is traversed from \((b,c)\) to \((b,d)\) and the left side from \((a,d)\) to \((a,c)\). Since \(dx=0\) on both sides, their total contribution is \[ \begin{aligned} &\int_c^d Q(b,y)\,dy + \int_d^c Q(a,y)\,dy\\ &\qquad= \int_c^d\bigl(Q(b,y)-Q(a,y)\bigr)\,dy\\ &\qquad= \int_c^d\int_a^b Q_x(x,y)\,dx\,dy, \end{aligned} \] by the fundamental theorem of calculus in the \(x\)-variable. This is \[ \iint_RQ_x\,dA. \]
4. Green’s theorem in step 1 holds.
Adding the horizontal contribution from step 2 and the vertical contribution from step 3 gives \[ \int_{\partial R}P\,dx+Q\,dy = \iint_R(Q_x-P_y)\,dA. \] This is the asserted formula.
5. Let\[ f=u+iv \]be analytic on an open neighborhood of \(R\). Then\[ u_x=v_y, \qquad u_y=-v_x \]on \(R\).
The real and imaginary parts of an analytic function satisfy the Cauchy–Riemann equations. In particular, the first partial derivatives used below are continuous on the rectangle, so Green’s theorem applies.
6. The real part of\[ \int_{\partial R}f(z)\,dz \]is zero.
5. Therefore\[ \operatorname{Re} \int_{\partial R}f(z)\,dz = 0. \]
7. The imaginary part of\[ \int_{\partial R}f(z)\,dz \]is zero.
5. Hence\[ \operatorname{Im} \int_{\partial R}f(z)\,dz = 0. \]
8. Cauchy’s theorem for a rectangle holds:\[ \boxed{ \int_{\partial R}f(z)\,dz=0. } \]
Steps <1>6–<1>7 show that both the real and imaginary parts of the complex integral vanish.
9. Q.E.D.
Steps <1>1–<1>4 state and prove Green’s theorem for rectangles, and steps <1>5–<1>8 use it to prove Cauchy’s theorem on a rectangle.