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
Integral over the boundary of a rectangle
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0sorry, i made the statement more clear – 2012-06-03
1 Answers
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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0ok, i got it. Thanks – 2012-06-04