Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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
up vote 1 down vote accepted

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 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 something similar for 2nd and 4th integrals.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.