Example, it's know

$\arcsin(x) = -i \ln ( \sqrt{1-x^2} + ix )$

All inverse trig functions have a log form, but when it's useful to write $-i \ln ( \sqrt{1-x^2} + ix )\:$ instead of $\:\arcsin(x)$ ? Is there any application or field (math, physics, eng etc.) when it's chosen the logarithm form?

It would be nice if someone can improve my tags because I don't know the right place to ask this question.

  • $\begingroup$ One possibility could be that if you were doing numerical calculations involving $\arcsin x$ for complex values of $x$, using the natural logarithm would be faster. $\endgroup$ – bjcolby15 Feb 25 '18 at 19:40

There are many applications related to quadratic transformations of hypergeometric ${}_2 F_1$ or ${}_3 F_2$ functions. For instance, $\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$ has an elementary proof due to the connection between $\arctan$ and $\log$: $$ \frac{\pi^2}{8}=\left[\frac{1}{2}\arctan^2 x\right]_{0}^{+\infty}=\int_{0}^{+\infty}\frac{\arctan x}{1+x^2}\,dx=\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+a^2 x^2)}\,dx\tag{1} $$ $$ \int_{0}^{1}\frac{-\log(a)}{1-a^2}\,da = \sum_{n\geq 0}\int_{0}^{1}-\log(a)a^{2n}\,da = \sum_{n\geq 0}\frac{1}{(2n+1)^2} = \zeta(2)-\sum_{\substack{n\geq 1\\n\text{ even}}}\frac{1}{n^2}=\frac{3}{4}\zeta(2).\tag{2}$$

Such connection can also be exploited to prove $$ \sum_{n\geq 1}\arctan\frac{1}{n^2}=\frac{\pi}{4}-\arctan\left(\frac{\tanh\frac{\pi}{\sqrt{2}}}{\tan\frac{\pi}{\sqrt{2}}}\right).\tag{3} $$ Indeed $\arctan\frac{1}{n^2}$ is the argument (i.e. $\text{Im}\log$) of $1+\frac{i}{n^2}$ and the infinite product $\prod_{n\geq 1}\left(1+\frac{i}{n^2}\right)$ can be computed through the Weierstrass product for the (hyperbolic) sine function.

Additionally, to write $\arcsin$ as a $\log$ is an effective way for proving the mysterious identity $$ \forall x\in(-1,1),\qquad \arcsin(x)^2 = \sum_{n\geq 1}\frac{4^n x^{2n}}{2n^2\binom{2n}{n}}\tag{4} $$ through Lagrange's inversion formula, see my notes.


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