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$\operatorname{Log}$ denotes the principal branch of the complex natural log

$\sqrt{t}$ denotes the principal square root of $t$

An antiderivative of $\sqrt{1-z^{2}}$ is $\frac{i}{2}(z\sqrt{z^{2}-1}-\operatorname{Log}(z+\sqrt{z^{2}-1})+C$

Is this statement correct?

I have of course differentiated it to check my work and it does seem that it's correct. However, what I'm not entirely sure about is choosing the principal branch of the natural log.

Wolfram does seem to confirm it, though I'm not sure if Wolfram has been careful enough about branches

Simon M
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  • Differentiate the result – Claude Leibovici Oct 06 '23 at 05:59
  • If I stick to using definite integrals, principal values, and simply connected domains, then I can unambiguously state that for all $z\in\mathbb{C}\setminus\mathbb{R}$ we have $\int_z^1 \sqrt{1 - z^2}\mathrm{d}z = \frac{\mathrm{i}}{2}(\mathrm{sgn},\Im z)(z^2\sqrt{1-z^{-2}}-\mathrm{arcosh},z)$. But what do your symbols mean? – K B Dave Oct 06 '23 at 07:09

2 Answers2

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Two different branches of $\operatorname{acosh}(z)$ differ by a constant. So if you use some other branch, you get the same answer, with a different constant $C$. Thus, for this integral, choosing the principal value is not a problem.

Of course it is good to be aware of this choice. In some other case, choice of the branch of a complex function may make a difference.

Explanation: $\cosh(z)$ is periodic with period $2\pi i$. If $\operatorname{acosh}(z)$ is the principal value of the inverse hyperbolic cosine, then the other values are $$ \operatorname{acosh}_k(z) := \operatorname{acosh}(z) + i 2 \pi k,\quad k \in \mathbb Z $$

Added. This is not quite right. These branches have derivative $\frac{1}{\sqrt{1-z^2}}$. There are other branches with derivative $-\frac{1}{\sqrt{1-z^2}}$. Such as $\frac{\pi i}{2}-\operatorname{acosh}(z)$.

GEdgar
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$\DeclareMathOperator{\arcosh}{arcosh}$ $\DeclareMathOperator{\sgn}{sgn}$ $\newcommand{\d}{\mathrm{d}}$ $\newcommand{\i}{\mathrm{i}}$

As an identity among principal branches of the functions involved (including the square root), it's wrong. What's true is that, if $\Re z \neq 0$ and $\Im z \neq 0$, then $$\int_1^z \sqrt{1-z^2}\d z = -\tfrac{\i}{2}(\sgn \Im z)((\sgn \Re z)z\sqrt{z^2-1}-\arcosh z)$$ so that your expression constitutes a valid indefinite integral only if $\Re z > 0$ and $\Im z<0$ (check a different quadrant numerically). This, even though it is true that $$\sinh \theta\,\d(\cosh \theta) =\tfrac{1}{2}\d(\cosh \theta \sinh \theta - \theta)$$ with $\theta = \mathrm{Arcosh}\,z$ a multivalued function of $z$. The lesson is that, when reducing identities involving multivalued functions to a single branch, one is not entitled to pick branches arbitrarily—they must be chosen coherently, so that the identities actually hold, and different choices might be required in different parts of the complex plane.

K B Dave
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  • What error could Wolfram be making when it tells me that the derivative of $\frac{i}{2}\left(z\sqrt{z^{2}-1}-\operatorname{arcosh}\left(z\right)\right)$ is $\sqrt{1-z^2}$? https://www.wolframalpha.com/input?i2d=true&i=derivative%20Divide%5Bi%2C2%5D%5C%2840%29zSqrt%5BPower%5Bz%2C2%5D-1%5D-arcosh%5C%2840%29z%5C%2841%29%5C%2841%29 – Simon M Oct 06 '23 at 23:48
  • @SimonM your premise is wrong. The expression in that link simplifies to $\sqrt{1-z^2}$ if and only if $\Re z > 0$ and $\Im z < 0$. It simplifies to something different in the other three quadrants. – K B Dave Oct 07 '23 at 00:42
  • Ok that also makes clear what you were getting at about the other quadrants. I understand now – Simon M Oct 07 '23 at 00:45
  • What is the method by which I can determine under what conditions the following is true? $\int\sqrt{1-z^{2}}dz=\frac{i}{2}(z\sqrt{z^{2}-1}-\operatorname{Log}(z+\sqrt{z^{2}-1})+C$ I really want to learn the method so that I don't make this same mistake again. – Simon M Oct 07 '23 at 00:56