Well-definedness of complex $\sin^{-1}$ I am define the inverse of complex $\sin$. I know that $\sin$ the conformal map mapping the strip $\{(x,y):x\in(-\pi/2,\pi/2), y>0\}$ to the upper half plane as the composition of the conformal maps $e^{iz}$ and $\frac{z-1/z}{2i}$. I am trying to find the inverse. I think $\int_0^z \frac{1}{{(1-s^2})^{1/2}}$ should play the role where the integral is over any contour from 0 to z and the branch for the square is $[-\pi/2,3\pi/2)$. Since $\sin^{-1}x=\int_0^x \frac{1}{{(1-s^2})^{1/2}}$
when $x\in[-1,1]$.
I am trying to use the analytic continuation. But I encountered some problems:

*

*$sin$ is the conformal map between the strip and the upper half plane, so its existence is guaranteed only on the upper half plane but not the real line. So I cannot use the analytic continuation directly from the fact that $\sin^{-1}x=\int_0^x \frac{1}{{(1-s^2})^{1/2}}$
when $x\in[-1,1]$.

*The behaviour of $\int_0^z \frac{1}{{(1-s^2})^{1/2}}$ on the real line. I think $\int_0^z \frac{1}{{(1-s^2})^{1/2}}$ is well-defined at $+1,-1$ even $\frac{1}{{(1-s^2})^{1/2}}$ has singularity at $+1,-1$, since $\int_0^1 1/s^{\beta} <\infty$ whenever $\beta<1$. But I do not know how to write down the argument for $\int_0^1 |\frac{1}{{(1-s^2})^{1/2}}|$ is finite on any contour connecting $0$ and $1$. Or more fundamentally, I do not know how to write down the rigorous argument for $\int_a^b |\frac{1}{z(t)^{1/2}}z'(t)|\le M\int_a^b |\frac{1}{z(t)^{1/2}}|<\infty$ where $z$ is the a contour from $0$ to $z$ in the upper half plane

 A: Let $a$ be the inverse of the restriction of $\sin$ to the whole vertical strip$$\left\{z\in\Bbb C\,\middle|\,\operatorname{Re}(z)\in\left(-\frac\pi2,\frac\pi2\right)\right\}.$$ Then\begin{align}a'(z)&=\left(\sin^{-1}\right)'(z)\\&=\frac1{\cos(a(z))}\end{align}and therefore\begin{align}\left(a'(z)\right)^2&=\frac1{\cos^2(a(z))}\\&=\frac1{1-\sin^2(a(z))}\\&=\frac1{1-z^2}.\end{align}Now, let $h(z)=\frac1{(1-z^2)^{1/2}}$, where, for each $z\in\Bbb C\setminus(-\infty,0]$, $z^{1/2}$ is the only square root of $z$ such that $\operatorname{Re}\left(z^{1/2}\right)>0$. It follows from what was done above that $(a')^2=h^2$ and that therefore $(a'-h)(a'+h)=0$. So, since we are dealing with analytic functions here, $a'=h$ or $a'=-h$. But we cannot have $a'=-h$, since $a'(0)=h'(0)=1$. So, $a'=h$ and therefore $a$ is the only primitive of $h$ which maps $0$ into $0$. In other words,$$a(z)=\int_{[0,z]}\frac1{(1-z^2)^{1/2}}\,\mathrm dz,$$where $[0,z]\colon[0,1]\longrightarrow\Bbb C$ is the path defined by $[0,z](t)=tz$. Still in other words,$$a(z)=\int_0^1\frac z{(1-z^2t^2)^{1/2}}\,\mathrm dt.$$
