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Let $G:=\mathbb{C}-\lbrace t\in \mathbb{R}\vert$ $\vert t\vert\geq1\rbrace$. Shows there is a representation for all $w\in G$ $$\arcsin(w)=\int_{\gamma(w)}\frac{1}{\sqrt{1-\zeta^{2}}}d\zeta$$ $\sqrt{1-\zeta^{2}}$ take the branch of root such that $(1-\zeta^{2})$ on $G$ assumes the value 1 on $\zeta=0$ an $\gamma(w)$ is an integration path form 0 to w on $G$.

I don't know what it means, simply shows that represntation is well defined or the formula $\arcsin(w)=\frac{1}{i}\mathrm{Log}(iw+\sqrt{1-w^{2}})$ and the representation integral are equivalent.

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Note that the only singularities of $$ \frac1{1-z^2}=\frac12\left(\frac1{1-z}+\frac1{1+z}\right)\tag{1} $$ are at $z=1$ and $z=-1$. If $\gamma$ is any closed path that does not circle these singularities, then by Cauchy's Integral Theorem $$ \int_\gamma\frac1{1-z^2}\mathrm{d}z=0\tag{2} $$ No path in $G$ can circle either singularity.

Since the difference of two paths between $0$ and $\zeta$ is a closed path, if both paths are in $G$, the difference of the integrals along those paths is $0$. That is, the function $$ \frac12\log\left(\frac{1+\zeta}{1-\zeta}\right)=\int_0^\zeta\frac1{1-z^2}\mathrm{d}z\tag{3} $$ is well-defined (it doesn't matter what path in $G$ is taken).

Exponentiating $(3)$ gives us a well-defined function $$ \frac1{1+\zeta}\exp\left(\frac12\log\left(\frac{1+\zeta}{1-\zeta}\right)\right)=\frac1{\sqrt{1-\zeta^2}}\tag{4} $$ which has no poles in $G$. Since $G$ is simply connected and there are no singularities of $\frac1{\sqrt{1-\zeta^2}}$ in $G$, we get a well-defined $$ \arcsin(w)=\int_0^w\frac1{\sqrt{1-\zeta^2}}\mathrm{d}\zeta\tag{5} $$

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  • $\begingroup$ why does this follows $\frac1{1+\zeta}\exp\left(\frac12\log\left(\frac{1+\zeta}{1-\zeta}\right)\right)=\frac1{\sqrt{1-\zeta^2}}$ $\endgroup$ – Porufes Sep 30 '14 at 3:07
  • $\begingroup$ @Porufes: Since we have a well-defined $\log\left(\frac{1+\zeta}{1-\zeta}\right)$, we exponentiate to get $\exp\left(\frac12\log\left(\frac{1+\zeta}{1-\zeta}\right)\right)=\sqrt{\frac{1+\zeta}{1-\zeta}}$ then divide by $1+\zeta$. $\endgroup$ – robjohn Sep 30 '14 at 5:23
  • $\begingroup$ oh and then you say that $\frac{1}{i}\mathrm{Log}(iw+\sqrt{1-w^{2}}) = \int_0^w \frac1{1+\zeta}\exp\left(\frac12\log\left(\frac{1+\zeta}{1-\zeta}\right)\right) \mathrm{d}\zeta $ ?? $\endgroup$ – Porufes Sep 30 '14 at 5:31
  • $\begingroup$ @Porufes: The integral is well defined. There are two functions that are not well-defined on $\mathbb{C}$ in $\frac1i\color{#C00000}{\log}\left(iw+\color{#C00000}{ \sqrt{\color{#000000}{1-w^2}}}\right)$: $\log(z)$ and $\sqrt{z}$. How are they supposed to be defined? $\endgroup$ – robjohn Sep 30 '14 at 9:57
  • $\begingroup$ I need to fix the branch were that make sense? $\endgroup$ – Porufes Sep 30 '14 at 10:49

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