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I was looking at the integral of $\frac{1}{\sqrt{x^2-1}}$ which equals $\text{arccosh}(x)$. This seemed very similar to $\int\frac{1}{\sqrt{1-x^2}} = \arcsin(x)$, which got me thinking about the following: $$ \begin{align} \text{arccosh}(x) &= \int\frac{1}{\sqrt{x^2-1}} \\ &= i \int\frac{1}{i\sqrt{x^2-1}} \text{ (as i is a constant)}\\ &= i \int\frac{1}{\sqrt{i^2 (x^2-1)}} \text{ (bringing i into the root)} \\ &= i \int\frac{1}{\sqrt{-1 (x^2-1)}} \text{ (as }i^2 = -1)\\ &= i \int\frac{1}{\sqrt{1-x^2}} \\ &= i \arcsin(x) \end{align} $$

The result $\text{arccosh}(x) = i \arcsin(x)$ seems absurd (and I could not find mention of this supposed equivalence anywhere). But according to this post, $i$ can be factored in and out of integrals, so where does my "proof" fail?

Henrik supports the community
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Suda
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1 Answers1

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$\newcommand{\arcosh}{\operatorname{arcosh}}\newcommand{\d}{\,\mathrm{d}}$You forgot the constant of integration: $$\arcosh(x)=i\arcsin(x)+C=i\arccos(x)+C$$

Since $\arcsin(x)=\frac{\pi}{2}-\arccos(x)$. Is this believable (with $C=0$)? Well, yes. It is true that $\cosh(iy)=\cos(y)$ and applying $\cosh$ to both sides finds: $$x=\cosh(\arcosh(x))\overset{?}{=}\cosh(i\arccos(x))=\cos(\arccos(x))=x$$So this is indeed reasonable as long as you allow for non-real interpretations of $\arccos,\arcosh$ (switching to a suitable complex definition of these two). For instance we can define: $$\arccos(z):=-i\log(z+\sqrt{z^2-1})$$For all complex $z$ and using the principal logarithm and principal square root. This definition agrees with the usual arccosine on $[-1,1]$ and if evaluated at $x\in[1,\infty)$ it would be literally true that $\arcosh(x)=i\arccos(x)$.

In fact, the usual, real-valued, principal $\arcosh$ is only defined on $[1,\infty)$ whereas the usual, real-valued, principal $\arccos$ is only defined on $[-1,1]$. So you're going to have to break some traditions to make this work.

The problem with the branches of these inverses arises when you casually decide to integrate $\int\frac{1}{\sqrt{1-x^2}}\d x$ where once you were integrating $\int\frac{1}{\sqrt{x^2-1}}\d x$. Really, integrals need bounds.

More accurately it is true that (for the principal real valued $\arcosh$): $$\arcosh(x)=\int_1^x\frac{1}{\sqrt{t^2-1}}\d t$$And when you move to: $$\cdots=i\int_1^x\frac{1}{\sqrt{1-t^2}}\d t$$There is an "issue" in that $\sqrt{1-t^2}$ is not a real number on the interval $1\le t\le x$. This is why we end up with non-standard inverses. But it is nonetheless true that $\cosh^{-1}(\{x\})=i\cos^{-1}(\{x\})$ for any $x$ as an equality of sets in $\Bbb C$ and this "issue" is no longer an issue if you choose holomorphic definitions of $\arccos$ on a suitable subset of $\Bbb C$.

FShrike
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