Why is there a difference between integrating it over a unit circle parametrized over $t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $t \in \left[-\frac{\pi}{2}, \frac{3\pi}{2}\right]$?
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Is the second interval perhaps meant to be $[\pi/2,3\pi/2]$? – Scounged May 15 '16 at 00:04
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1It would perhaps help if you wrote the integral out in full – Alexander McFarlane May 15 '16 at 00:22
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In agreement with @AlexanderMcFarlane, I say that your question can not be answered till you show us the parametrization you used. It seems to me that if you are integrating a continuous function, the value of your function will be different at the two endpoints. – Lubin May 15 '16 at 00:54
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@Scounged: yes! It should be, I copied and pasted out of laziness and forgot to remove the minus. – Edi Madi May 15 '16 at 01:11
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The difference comes from $e^{i\theta}$ giving different signs of $\pm i$ for the different angles.
The first approach I would always take on these problems is by looking for branch cuts in the complex logarithm by doing the sneaky move of replacing as,
$$ z^i = e^{i\ln{z}} = e^{i\ln{re^{i\theta}}} $$ But $r=1$ on the unit circle,
$$ e^{i\ln{e^{i\theta}}} $$
Now at this point I'll link to an explanation on branch cuts (by someone related to me!) with relevance to the complex phase $e^{i\phi}$, which should give you an awareness of some of the concepts as you have a complex phase with a phase angle of $\phi=\ln{e^{i\theta}}$
Also note that $\ln(\pm i)=\pm \frac{i\pi}{2}$
As it turned out your phase wasn't actually even complex as all the values you provide give a nice $\phi = \ln{(\pm i)}$ so you will obtain something like $e^{\mp \pi /2}$ where the difference is due to my opening assertion
Alexander McFarlane
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Yes I'm just editing now I'm typing it on my iPhone though so inserted the $i$ one character early! – Alexander McFarlane May 14 '16 at 23:44