Take the tour ×  Ok so to back up a bit, by a trig substitution
we have for $f:(-1,1)\rightarrow\mathbb{R}$:
$$f(x)=\int_0^x\frac{dt}{\sqrt{1-t^2}} = \arcsin(x)$$  Now
according to the notes here: Elliptic Functions Notes, the proof
that for a 'large' circle centered at zero, the fact that
implies the periodicity of the sine function. Can anyone else
explain to me what's going on here? I'm really lost. I mean
arcsine is of course multivalued unless we choose a particular
'branch' or whatever, but I thought we chose this 'branch'
implicitly when we did our trig-sub to solve the integral in the
first place. I suppose by analytic continuation, this integral
must be equal to $\arcsin(z)$ on $\mathbb{C}-\{-1,1\}$. But how
they go from there to talking about the periodicity of sine
based just off this integral's value along a closed path
containing both poles I simply don't understand. Thanks.
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