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I discovered this relation while playing around with $\int\frac{\mathrm dx}{\sqrt{x^2+1}}$.

Method 1:

This is a standard integral, so it can be directly written as: $$\int\frac{\mathrm dx}{\sqrt{x^2+1}}=\sinh^{-1}x+C$$

Method 2:

The integrand can be rewritten in the form $\frac1{\sqrt{1-t^2}}$ if $t=\pm ix\left(i=\sqrt{-1}\right)$. For simplicity, I took $t=ix$. Here, I have made the assumption that the domain of the function $\sin^{-1}x$ can be extended to $\mathbb C$:

$$-i\int\frac{\mathrm d(ix)}{\sqrt{1-(ix)^2}}=-i\sin^{-1}ix+C$$

So, $\sinh^{-1}x$ and $-i\sin^{-1}ix$ differ by a constant: $$\sinh^{-1}x=-i\sin^{-1}ix+C$$

To find $C$, I put $x=0$ in the above equation: $$0=0+C$$ $$\therefore \sinh^{-1}x=-i\sin^{-1}ix$$

But since this did not seem like a rigorous proof of the formula, I tried to use the relation $\sinh x=-i\sin{ix}$ to prove it:

$$\text{Let }y=\sinh^{-1}x\implies x=\sinh y \text{ as} \sinh x \text{ is a bijective function }\forall x\in \mathbb R$$ $$\implies x=-i\sin iy$$ $$\implies ix=\sin iy$$ $$\implies \sin^{-1}ix=\sin^{-1}\sin iy$$

This is where I got stuck. I had no clue how to evaluate $\sin^{-1}\sin x$ for imaginary values of $x$. I know how to do it for real $x$, but I am not familiar with working with inverse trigonometric functions when their domain is extended to $\mathbb C$. Can someone suggest me how to proceed forward? Or is there any other way to prove this?

I tried to look up my query on this site, but could not find any relevant post. If someone does find such a post, kindly share it by commenting before downvoting without any explanation. I was able to find these links on Quora, but both of them implicitly used $\sin^{-1}\sin ix=ix, x\in\mathbb R$.

Integreek
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    probably need to start from definition 2 in here https://proofwiki.org/wiki/Definition:Inverse_Sine/Complex, and work out if you can constraint to one branch only for your particular integral. – zwim Sep 21 '24 at 11:31
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    Do you know that $i\sin(z) = \sinh(iz)$ for all complex numbers $z$? – Martin R Sep 21 '24 at 11:40
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    Those who have doubts or require any clarification, please comment before downvoting. – Integreek Sep 21 '24 at 12:05
  • @MartinR yes I do, it directly comes from $\sinh x=-i\sin ix=\frac{e^x-e^{-x}}2$. I am familiar with taking sine, cosine,etc. of complex numbers since the trig ratios of imaginary numbers obtained on applying compound angle formulae can be expressed via hyperbolic functions. – Integreek Sep 21 '24 at 14:32
  • @zwim thanks, that helped me complete the proof. But where did this formula come from? Or is it simply defined that way? – Integreek Sep 21 '24 at 14:44
  • Why the downvote? Please explain. – Integreek Oct 08 '24 at 07:56
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    Please cut down on minor edits, @MathGuy. – Shaun Oct 08 '24 at 09:05

1 Answers1

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Let

$$ \sinh^{−1}x = y, \tag{1} $$

and

$$ -i\sin^{-1}ix = z. \tag{2} $$

Now I will try to prove $y = z$.

From (2),

$$ \Rightarrow \sin^{-1}ix=-\frac{z}{i} \\ \Rightarrow ix=\sin(-\frac{z}{i})=\frac{e^{-z}-e^z}{2i} \\ \Rightarrow x=\frac{e^{z}-e^{-z}}{2}=\sinh z \\ \Rightarrow z=\sinh^{-1}x=y. $$

Thus proved.

Miscellaneous
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  • Does $\sin(\sin^{-1} x)=x$ hold for any imaginary value of $x$? Is it obtained by extrapolating this relation for $x\in [-1,1]$? – Integreek Sep 25 '24 at 03:53
  • Or, in other words, is $\sin^{-1}x$ a bijective function from $\mathbb C-\mathbb R\to \mathbb C-\mathbb R$? – Integreek Sep 25 '24 at 04:03
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    Could be, if the branch cut is selected correctly. If the branch cuts for inverse $\sinh$ and $\sin$ are both selected to be the same , say $[-\pi, \pi]$, then everything is fine here. – Miscellaneous Sep 25 '24 at 04:09
  • Can you please explain branch cut in simple terms? On the internet, it is explained using the terms singularities, analytic function which I am not able to understand since I am in high school. – Integreek Sep 25 '24 at 04:15
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    Without branch cut, $\sin^{-1}x$ is multi-valued. Say $x=1$, then $\sin^{-1}x=-\pi/2, \pi/2, 3\pi/2, \cdots $. Selecting a branch $(-\pi, \pi]$ means to accept only values in the range $(-\pi, \pi]$ as the function value of $\sin^{-1}x$. – Miscellaneous Sep 25 '24 at 04:22
  • Ok thank you, I got it! – Integreek Sep 27 '24 at 16:21
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    @MathGuy The branch cut also explains why there is an extra constant $C$ in your method 1. If you don't choose a branch cut, with the same example $x=1$, $\sin^{−1}x=−π/2,π/2,3π/2,\cdots=-\pi/2 + n\pi$, the term $n\pi$ here is the integration constant $C$. – Miscellaneous Sep 28 '24 at 02:14