Arc length of an arc of semi-circle diverges

Question

Given the unit circle $S^1=\{(x,y)\in\mathbb{R^2}\mid x^2+y^2=1\}$, We know that its length (As a closed simple curve) is $2\pi$ and in fact finite.

Suppose we take only its positive half, defined by the function $f(t)=\sqrt{1-t^2}$. One may assume that $$\ell(f[0,1])=\frac{2\pi}{4}=\frac\pi 2.$$

However, $$\ell(f[0,1])=\int_0^1\sqrt{1+f^\prime(t)^2}dt=\int_0^1\sqrt{1+\frac{t^2}{1-t^2}}dt = \int_0^1\frac{1}{\sqrt{1-t^2}}dt$$ which diverges.

Something seems fallacious with this argument but I can't point on it. What's wrong with it?

Answer 1

$$\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}dx=\arcsin(1)-\arcsin(0)=\frac{\pi}{2}$$ other way $$x=r\cos\theta\quad,\quad y=r\sin\theta\implies x^2+y^2=r^2=1$$ apply $$l(f[\theta_1,\theta_2])=\int_{\theta_1}^{\theta_2}\sqrt{r^2+r'\,^2}d\theta=\int_{0}^{\frac{\pi}{2}}\sqrt{1^2+0^2}d\theta=\frac{\pi}{2}$$

Answer 2

Observe that

$$\int_0^1\frac{dt}{\sqrt{1-t^2}}\stackrel{\text{Def.}}=\left.\lim_{\epsilon\to0^+}\arcsin t\right|_\epsilon^1=\arcsin 1-0=\arcsin1=\frac\pi2$$

and thus the improper integral converges.

In the above we take the usual definition intervals:

$$\sin x\;,\;\;x\in\left[-\frac\pi2,\,\frac\pi2\right]\;\;;\;\;\;\arcsin x\;,\;\;x\in\left[-1,\,1\right]$$