Guys I wonder if you can help me out with a certain integral, in this case proving convergence. The integral seems simple but there is a catch, I am using this integral to define the sine and cosine functions, so you can't use inverse sine and cosines just yet in order to prove convergence as that would lead to circular reasoning.
I am basically trying to define the sine and cosine in a rigorous way, by building on the high school definition of the unit circle but translating that into something rigorous. I basically translated the high school definition into this:
The Sin is the unique function satisfying the following definition:
$\sin(\int_{x}^{1} \frac{1}{\sqrt{1-t^2}} dt) = \sqrt{1-x^2}$
Where I get the integral by calculating the length of the arc of the half circle above the interval $[-1,1]$, in the standard way. The sweet thing is that I can now define pi as:
$\pi = \int_{-1}^{1} \frac{1}{\sqrt{1-t^2}}dt$
This basically mimics the high school definition if you take a better look, and therefore is more suitable in my opinion than a definition based on the taylor series or something like that, as it translates intuition into rigorous maths in a more direct way.
The problem I come across when defining this that the integral is not a proper one, as the function inside diverges if x goes to 1 or -1. So my basic question is, how could I prove the above integral actually converges(I don't need a value, a proof of convergence is just fine). One way would be to use another function that is strictly greater at every point and that does converge on the interval. However I couldn't find an example of such a function.
Another approach I thought about was reasoning along these lines:
Consider the function $f(x) = \sqrt{1-x^2}$, the arc length of this thing above $[-1,1]$(that I just defined as $\pi$, very naturally) can be calculated by:
$ L = \int_{-1}^{1} \sqrt{1+(f'(x))^2} dx = \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} dx$
Now we know $[-1,1]$ is a compact set, and f is continuous and that in turn implies that $f([-1,1])$ must be compact as well. So the graph $Gr(f) = \{ (x,f(x)) \in \Bbb R^2 | x \in [-1,1] \} $ is compact as well. But compactness doesn't necessarily imply a finite one dimensional measure in a 2D space, as fractals definitely provide a counterexample. The function given is infinitely many times differentiable so maybe that could be used.
I am not aware of any general theorem on the finiteness of arc length of functions whose graphs are compact, but to me it wouldn't seem far fetched if it did exist in some form. So something along the following lines:
Let $f: A $->$ B$, $A, B \subset \Bbb R$ be a function where both A and f(A) are compact sets, and let f satisfy some additional properties, then the integral $ \int_{A} \sqrt{1+(f'(x))^2} dx$ converges even if it isn't a proper integral.
It would be neat if such a theorem exists, but i'm not aware of it. Either of the 2 approaches should be able to prove convergence, and therefore the well definedness of the sine.
