Let $\gamma:[0,1] \longrightarrow \mathbf{C} \backslash \{0\}$ be a closed curve (continuous and of bounded variation). 
We call 
$$\operatorname{Ind}\_\gamma(0) \overset{\mathrm{def}}{=} \frac{1}{2 \pi i}\int\_\gamma \frac{1}{z}\ dz.$$
$\textbf{the index of $\gamma$ around $0$.}$
As for the winding number, we can take a pair of continuous real valued functions $r$ and $\theta$ that satisfies
$$r:[0,1] \longrightarrow (0,\infty),$$
$$\theta:[0,1] \longrightarrow \mathbf{R},$$
$$\forall t \in [0,1]\, \left[\, \gamma(t) = r(t) \cdot e^{i \cdot \theta(t)}\, \right],$$
and
$$-\pi < \theta(0) \leq \pi.$$
(The proof of this can be found in chapter 7 of A. F. Beardon, Complex Analysis: The Argument Principle in Analysis and Topology, Wiley-Interscience publication 1979).
We call
$$\operatorname{Wnd}\_\gamma(0) \overset{\mathrm{def}}{=}\frac{\theta(1) - \theta(0)}{2\pi}$$
$\textbf{the winding number of $\gamma$ around $0$}$.

It is already known that if $\gamma$ is $C^1$-curve, then
$$\operatorname{Ind}\_\gamma(0) = \operatorname{Wnd}\_\gamma(0)$$
(see [definition of winding number, have doubt in definition.](https://math.stackexchange.com/questions/186512/definition-of-winding-number-have-doubt-in-definition)).

My question: When $\gamma$ is not $C^1$ path,
$$\operatorname{Ind}\_\gamma(0) = \operatorname{Wnd}\_\gamma(0)?$$

Thank you for reading.
['https://math.stackexchange.com/questions/3257640', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/680871/']
First, we observe that
$$
f(x)=x-\ln x -1\geq \frac{(x-1)^2}{2}-\frac{(x-1)^3}{3}=g(x).
$$
Now clearly if 
$$
g(x)\geq\varepsilon \Rightarrow f(x)\geq \varepsilon
$$

Now all we have to do is to find the roots $f(x)-\varepsilon=0$ and identify the one which is larger than, but close to $1$.
It is given by
$$
x(\varepsilon)=-\frac{1}{4}(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}-\frac{1}{4(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}}+\frac{3}{2}-\frac{i\sqrt{3}}{2}\left(\frac{1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3}}{2}-\frac{1}{2(1-12\varepsilon+2\sqrt{36\varepsilon^2-6\varepsilon})^\tfrac{1}{3})}\right)
$$

edit: If this expression does not satisfy you, you can of, of course expand it around $\varepsilon=0$ to get:

$$
x(\varepsilon)\approx 1+\sqrt{2\varepsilon}+\tfrac{2}{3}\varepsilon+\tfrac{5}{9}\sqrt{2}\varepsilon^\tfrac{3}{2}
$$
$x(\epsilon)=x-\ln(x)-1$ is the largest upper bound possible, and also happens to be in closed form.