Prove that the integral $\int\frac{\tau(s)}{\kappa(s)}ds=0$ Let $\vec r=\vec r(s)$ be a closed regular curve on $S^2\subset\mathbb R^3$, parametrized by arclength. Prove that
$$\int\frac{\tau(s)}{\kappa(s)}ds=0$$
where $\tau$ and $\kappa$ are respectively the torsion and the curvature of the curve.
My progress: I think I am supposed to write $\tau/\kappa$ as a derivative something else. But the only information I can obtain from $\vec r$ lying on the unit sphere is that
$$
\vec r(s)=0\vec t-\frac{1}{\kappa}\vec n-\frac{1}{\tau}\frac{d}{ds}\Big(\frac{1}{\kappa}\Big)\vec b
$$
where $\vec t,\vec n,\vec b$ is the Frenet frame of $\vec r$. Since $\vec r$ has length $1$ I can write
$$\frac{\tau}{\kappa}=\pm\sqrt{1-\bigg(\frac{d}{ds}\Big(\frac{1}{\kappa}\Big)\bigg)^{\!2}}$$
which does not help me solve the integral.
Any suggestions?
 A: Let $\omega$ denote a period for the curve $\vec{r}$. It is useful to know that for a curve $\vec{r}$ lying on $S^{2}\subset\mathbb{E}^{3}$, the curvature $\kappa$ and torsion $\tau$ can be expressed in terms of the function 
$$
j(s):=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right).\tag{*}
$$
Namely, I claim that
\begin{align}
&\kappa(s)=\sqrt{1+j(s)^{2}},\tag{1}\\
&\tau(s)=\frac{j'(s)}{1+j(s)^{2}}\tag{2}.
\end{align}
Assuming for now that the equalities $(1)$ and $(2)$ are true, we get
$$
\int_{0}^{\omega}\frac{\tau(s)}{\kappa(s)}ds=\int_{0}^{\omega}\frac{j'(s)}{(1+j(s)^{2})^{3/2}}ds=\int_{j(0)}^{j(\omega)}\frac{1}{(1+u^{2})^{3/2}}du,\tag{3}
$$
substituting $u:=j(s)$ in the last equality. Now, since $\vec{r}(s)=\vec{r}(s+\omega)$, we know in particular that $\vec{r}(0)=\vec{r}(\omega)$, $\dot{\vec{r}}(0)=\dot{\vec{r}}(\omega)$ and $\ddot{\vec{r}}(0)=\ddot{\vec{r}}(\omega)$. So from $(*)$ it is then clear that $j(0)=j(\omega)$, which implies that the integral $(3)$ is indeed zero.
Now it remains to prove the equalities $(1)$ and $(2)$.
(1) Since $\vec{r}(s)$ lies on the unit sphere and has speed $1$, we have that $\{\vec{r}(s),\dot{\vec{r}}(s),\vec{r}(s)\times\dot{\vec{r}}(s)\}$ is an orthonormal frame. Expressing $\ddot{\vec{r}}(s)$ in this frame gives
\begin{align*}
\ddot{\vec{r}}(s)&=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+(\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s))\dot{\vec{r}}(s)+(\ddot{\vec{r}}(s)\cdot \vec{r}(s)\times\dot{\vec{r}}(s))\vec{r}(s)\times\dot{\vec{r}}(s)\\
&=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s)\\
&=-\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s).\tag{4}
\end{align*}
Here we use that $\vec{r}(s)\cdot\vec{r}(s)=1$, so that $\dot{\vec{r}}(s)\cdot\vec{r}(s)=0$ and therefore $\ddot{\vec{r}}(s)\cdot\vec{r}(s)=-\dot{\vec{r}}(s)\cdot\dot{\vec{r}(s)}=-1$. We also use that $\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=0$ since $\dot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=1$. Now finally, since $\kappa(s)=\|\ddot{\vec{r}}(s)\|$, the expression $(4)$ gives
$$
\kappa(s)=\|\ddot{\vec{r}}(s)\|=\sqrt{1+j(s)^{2}}.
$$
(2) We can rewrite the expression (*) as follows:
\begin{align*}
j(s)&=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\dot{\vec{r}}(s)\times\ddot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\vec{t}(s)\times\kappa(s)\vec{n}(s)\right)\\&=\kappa(s)\vec{r}(s)\cdot\vec{b}(s).
\end{align*}
Differentiating the equality $\vec{r}(s)\cdot\vec{r}(s)=1$ repeatedly and using the Frenet-Serret formulas, we get
\begin{align*}
\vec{r}(s)\cdot\vec{t}(s)=0 &\Rightarrow \vec{t}(s)\cdot\vec{t}(s)+\kappa(s)\vec{r}(s)\cdot\vec{n}(s)=0\\
&\Rightarrow \vec{r}(s)\cdot\vec{n}(s)=-1/\kappa(s).
\end{align*}
So, expressing $\vec{r}(s)$ in the orthonormal frame $\{\vec{t}(s),\vec{n}(s),\vec{b}(s)\}$ gives
\begin{align*}
\vec{r}(s)&=(\vec{r}(s)\cdot\vec{t}(s))\vec{t}(s)+(\vec{r}(s)\cdot\vec{n}(s))\vec{n}(s)+(\vec{r}(s)\cdot\vec{b}(s))\vec{b}(s)\\
&=\frac{-1}{\kappa(s)}\vec{n}(s)+\frac{j(s)}{\kappa(s)}\vec{b}(s).
\end{align*}
Differentiating this equality and using the Frenet-Serret formulas gives
$$
\vec{t}(s)=-\left(\frac{1}{\kappa(s)}\right)'\vec{n}(s)-\frac{1}{\kappa(s)}(-\kappa(s)\vec{t}(s)+\tau(s)\vec{b}(s))+\left(\frac{j(s)}{\kappa(s)}\right)'\vec{b}(s)-\frac{j(s)}{\kappa(s)}\tau(s)\vec{n}(s)\tag{5}
$$
So the coefficient of $\vec{b}(s)$ in $(5)$ has to be zero, which gives
$$
\tau(s)=\kappa(s)\left(\frac{j(s)}{\kappa(s)}\right)'.\tag{6}
$$
But we know from $(1)$ that $\kappa(s)=\sqrt{1+j(s)^{2}}$. So
$$
\left(\frac{j(s)}{\kappa(s)}\right)'=\frac{j'(s)\sqrt{1+j(s)^{2}}-\frac{j(s)^{2}j'(s)}{\sqrt{1+j(s)^{2}}}}{1+j(s)^{2}}=\frac{j'(s)(1+j(s)^{2})-j(s)^{2}j'(s)}{(1+j(s)^{2})^{3/2}}=\frac{j'(s)}{(1+j(s)^{2})^{3/2}}.\tag{7}
$$
So we conclude from $(6)$, $(7)$ and $(1)$ that
$$
\tau(s)=\frac{j'(s)}{1+j(s)^{2}}.
$$
A: This is a theorem of B. Segre from 1947. See reference 6. in the page visible here:
https://link.springer.com/article/10.1007/s00022-017-0397-8
