Talk:Arcsine as Integral/Lemma 1
Here is the proof as I have it now.
Proof
For this proof only, let $\sin_A$ be the analytic sine function from Definition:Sine/Complex Numbers.
Consider $\int_{0}^{x}\frac{1}{\sqrt{1-x^2} }\d{x}$
Let $x=\sin_A\left(\Theta\right)\iff x=\arcsin_A\left(\Theta\right)$
| \(\ds \d{x}\) | \(=\) | \(\ds \cos_A\left(\Theta\right)\d{\Theta}\) | Derivative of Sine Function | |||||||||||
| \(\ds \int\frac{1}{\sqrt{1-x^2} }\d{x}\) | \(=\) | \(\ds \int\frac{1}{\cos_A\left(\Theta\right)}\cos_A\left(\Theta\right)\d{\Theta}\) | Integration by Substitution | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int1d\Theta\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \Theta+C\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \arcsin_A\left(x\right)+C\) | ||||||||||||
| \(\ds \int_{0}^{x}\frac{1}{\sqrt{1-x^2} }dx\) | \(=\) | \(\ds \arcsin_A\left(x\right)\) | Fundamental Theorem of Calculus/Second Part |
$\Box$
I would like to change this proof so it reads as follows:
Proof
For this proof only, let $\sin_A$ be the analytic sine function from Definition:Sine/Complex Numbers.
| \(\ds \int_0^x \frac 1 {\sqrt{1 - t^2} } \rd t\) | \(=\) | \(\ds \arcsin x\) | Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form/Corollary 2 |
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But in establishing the above, only the analytic sine and Arc Sine, $\sin_A$, and $\arcsin_A$ should be used. This avoids circular reasoning in proving Derivative of Sine Function in the geometric case.
$\Box$
The problem is, that if I made this change, I am not sure that the reader would understand what I am saying and why. It may be the original version, even if longer and more cumbersome, may be better understood. I would welcome suggestions and help here.
- Immediate problem: the integration as presented here applies to the real case only. For this to apply on the complex case, you would need to set up a contour integral. Is it really necessary to invoke the complex definition? --prime mover (talk) 11:41, 29 January 2019 (EST)
For the purposes I have here, I only need the real version. But if I remember my complex analysis correctly, it does not matter what path you take so long as you stay in a compact region where the function is continuous. but if you go around the singularity at the origin, you can get different answers. Please check me, I might be wrong. I am not sure how to generalize to complexes. --Pelliott (talk) 16:51, 29 January 2019 (EST)
- True, but then of course first you have to demonstrate that the function in question is analytic, which IIRC (I may be wrong, long time since I cracked open my complex analysis books) you need to know its derivative in order to do so. Circularity again. But that's not the point. The notation you are using is that of the real integral. Complex integration has a completely different notation and interpretation. --prime mover (talk) 17:15, 29 January 2019 (EST)