Deriving $\int \frac{dz}{z\sqrt{1-z^2}}=\operatorname{Log}(\frac{\tanh(\frac{arcosh(z)}{2})+i}{\tanh(\frac{arcosh(z)}{2})-i})+C$

calculuscomplex-analysisindefinite-integralsintegrationsolution-verification

The principal branch is assumed for all mentioned trigonometric and inverse trigonometric functions

$\operatorname{Log}$ denotes the principal branch of the natural log defined over the complex numbers
$\DeclareMathOperator{\artanh}{artanh}$
$\DeclareMathOperator{\arctan}{arctan}$
$\DeclareMathOperator{\arcosh}{arcosh}$

First, we can show that
$$\int \frac{dz}{z\sqrt{1-z^2}}=\int \frac{\sinh\left(\theta\right)d\theta}{\cosh\left(\theta\right)\left(i\sinh\left(\theta\right)\right)}=-i\int\frac{d\theta}{\cosh\left(\theta\right)}=-2i\arctan\left(\tanh\left(\frac{\theta}{2}\right)\right)+C$$
$$=-2i\arctan(\tanh(\frac{\arcosh(z)}{2}))+C$$
We can also show that

$$\int \frac{du}{1+u^2}=\int \frac{i/2}{u+i}-\frac{i/2}{u-i}du=\frac{i}{2}\operatorname{Log}\left(\frac{u+i}{u-i}\right)+C_1=\arctan(u)+C_2$$

And so we obtain

$$\operatorname{Log}\left(\frac{u+i}{u-i}\right)=-2i\arctan\left(u\right)+C$$

Substitute
$$u=\tanh\left(\frac{\arcosh\left(z\right)}{2}\right)$$

Then using the above we can finally arrive at
$$\int \frac{dz}{z\sqrt{1-z^2}}=\operatorname{Log}\left(\frac{u+i}{u-i}\right)+C=\operatorname{Log}(\frac{\tanh(\frac{\arcosh(z)}{2})+i}{\tanh(\frac{\arcosh(z)}{2})-i})+C$$

$$=\operatorname{Log}\left(\frac{\frac{z+\sqrt{z^{2}-1}-1}{z+\sqrt{z^{2}-1}+1}+i}{\frac{z+\sqrt{z^{2}-1}-1}{z+\sqrt{z^{2}-1}+1}-i}\right)+C$$

Best Answer

$\DeclareMathOperator{\Arsech}{Arsech}$ $\DeclareMathOperator{\Arcosh}{Arcosh}$ $\DeclareMathOperator{\Arctan}{Arctan}$ $\DeclareMathOperator{\arsech}{arsech}$ $\DeclareMathOperator{\artanh}{artanh}$ $\DeclareMathOperator{\arcosh}{arcosh}$ $\DeclareMathOperator{\arcsec}{arcsec}$ $\DeclareMathOperator{\sgn}{sgn}$ $\newcommand{\d}{\mathrm{d}}$ $\newcommand{\i}{\mathrm{i}}$

I use common conventions for the inverse trigonometric functions, including using lowercase for the conventional principal branches and title case for the corresponding multivalued functions. (The square root $\sqrt{(-)}$ is also taken to be the conventional principal branch.) Then by definition your integral is equal to

$$-\Arsech z +C$$ and your question, in essence, is whether you have correctly demonstrated a relation between $-\Arsech z$ and the inverse tangent, hyperbolic tangent, and inverse hyperbolic cosine. Working with principal branches for the moment, I think your derivation can be usefully interpreted and decomposed into three true claims:

  1. If $z\in \mathbb{C}\setminus\mathbb{R}$, then $$-\arsech z = \i(\sgn\Im z)\arcsec z\text{.}$$ This is an implicit analytic continuation you made that introduced the $\i$ in the hyperbolic substitution.
  2. If $z\in \mathbb{C}\setminus (-\infty,1]$, then $$\arcosh z = 2 \artanh \sqrt{\frac{z-1}{z+1}}\text{.}$$ This is the hyperbolic substitution that you made.
  3. If $z\in \mathbb{C}^{\star}\setminus [-1,1]$, then $$\arcsec z = 2 \arctan \sqrt{\frac{z-1}{z+1}}\text{.}$$ This is the evaluation of the integral in terms of the inverse tangent (as a Weierstrass substitution).

Therefore, it is true that, if $z\in \mathbb{C}\setminus \mathbb{R}$, then

$$\boxed{-\arsech z = 2\i(\sgn\Im z)\arctan\left(\tanh (\tfrac{1}{2}\arcosh z)\right)}\text{.}$$

The multivalued versions of the inverse functions appearing are $$\begin{align} \Arsech z &=\pm\arsech z + 2\pi \i\mathbb{Z} \\ \Arcosh z &=\pm\arcosh z + 2\pi \i\mathbb{Z} \\ \Arctan z &=\arctan z + \pi\mathbb{Z}\text{.} \end{align}$$ Therefore, it is true that, as multivalued functions, $$\boxed{-\Arsech z = 2\i\Arctan\left(\tanh(\tfrac{1}{2}\Arcosh z)\right)}\text{.}$$

Note, however, that it is crucial that both inverse functions appearing on the right-hand side be multiply valued—we lose branches otherwise: $$\begin{align}2\i\arctan\left(\tanh(\tfrac{1}{2}\Arcosh z)\right)&=\pm\arsech z \\ 2\i\Arctan\left(\tanh(\tfrac{1}{2}\arcosh z)\right) &= -\arsech z + 2\pi\i\mathbb{Z}\text{.} \end{align}$$

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