# Integral over the boundary of a rectangle

This might be astupid question, but i really got stuck in it! So basically i have $F:\mathbb{C} \longrightarrow \mathbb{C}$ meromorphic, $\tau$ a complex number such that $Im(\tau)>0$and $0<Re(\tau)<1$, and F has the periodicity $F(z+1)=F(z)=F(z+\tau)$. Suppose there are no zeros nor poles in the boundary of the rectangle whose vertices are $0,1,1+\tau,\tau$. I have to integrate $F'/F$ on the boundary of this retangle, and this should be zero. Parametrizing the segments in this way: $\gamma_1(t)=t$, $\gamma_2(t)=1+t\tau$,$\gamma_3(t)=1-t+\tau$,$\gamma_4(t)=(1-t)\tau$ (they all have domain $[0,1]$), and using the periodicity I obtain $$\int_0^1\frac{F'(t)}{F(t)}dt+\int_0^1\tau\frac{F'(t\tau)}{F(t\tau)}dt-\int_0^1\frac{F'(-t)}{F(-t)}dt-\int_0^1\tau\frac{F'(-t\tau)}{F(-t\tau)}dt$$. This should be zero because but i cannot explain to myself why he first and the third integral are opposite (and so the 2 and the 4).

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By "vertical half semispace" do you mean the y-axis, or if you will the imaginary one? Because otherwise I can't see how $\,0,1,1+\tau,\tau\,$ can be the vertices of a rectangle... –  DonAntonio Jun 3 '12 at 21:59
sorry, i made the statement more clear –  balestrav Jun 3 '12 at 22:04

Why would you do such a messy parametrization of the rectangle (which is not a rectangle unless $\,\operatorname{Re}\tau=0\,$)? This looks as the beginning of elliptic functions and stuff, when you show how integrals of such functions as $\,F\,$ vanish on a fundamental paralleliped on the upper complex plane (and thus in fact $\,\operatorname{Im}\tau>0\,$) ..., but use periodicity of $\,F\,$ ! This function is the same on opposite sides of this parallelogram, which your integral "walks" on in opposite directions, so their values cancelate each other...as simple as that!

*Added*$\,\,\,$ Taking your parametrizations of the four segments, we get: $$\int_0^1\frac{F'(t)}{F(t)}dt+\tau\int_0^1\frac{F'(1+t\tau)}{F(1+t\tau)}dt-\int_0^1\frac{F'(1-t)}{F(1-t)}dt-\tau\int_0^1\frac{F'((1-t)\tau)}{F((1-t)\tau)}dt$$Put now $\,\,u=1-t\Longrightarrow du=-dt\,$ in the 3rd integral to get $$\int_1^0\frac{F'(u)}{F(u)}(-du)=\int_0^1\frac{F'(u)}{F(u)}du$$which is exactly your first integral, and now use that minus sign before the 3rd integral...do something similar for 2nd and 4th integrals.

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It is in fact, but i wanted to prove it by doing the integral from the definition.. is there any way to go on from my calculations rigorously? –  balestrav Jun 3 '12 at 22:10
I guess so, but I see you forgot completely in all your line integrals converted to "usual" parametrized integrals to multiply by the differential of the substitution! For example, in $\gamma_2(t):=1+t\tau\,$, we obtain $\,\gamma_2'(t)=\tau\,dt\,$...check and correct this –  DonAntonio Jun 3 '12 at 22:14
Yes, I corrected it and it shuold be right now, but i have still the same problem! –  balestrav Jun 3 '12 at 22:20
@Balestrav Why did you write your second, third and fourth integrals as you did? The 2nd one, for ex., should be $\,\displaystyle{\int_0^1\tau\,dt\frac{F'(1+t\tau)}{F(1+t\tau)}}\,$ , and something similar with the 4th one, and the 3rd one is even simpler if you leave first the argument as $\,1-t\,$ and then check the signs... –  DonAntonio Jun 3 '12 at 22:25
great! Thanks a lot! Only one doubt: how did I use the periodicity in this way? It seems that only a change of variables is required.. –  balestrav Jun 3 '12 at 22:33