Integrate logarithmic derivative of a periodic function Given $f$ a $p$-periodic function over $\mathbb{C}$, how to show that :
$$\frac{1}{\mathrm{i}p}\int_a^{a+p}\frac{f'(t)}{f(t)}dt \in \mathbb{Z}$$
Is there any elegant method ?
 A: Let $\gamma: [0, 1] \to \Bbb C$ be a path connecting $a$ and $a+p$.
Then 
$$
 \int_a^{a+p}\frac{f'(z)}{f(z)} \, dz = \int_0^1 \frac{f'(\gamma(t))\gamma'(t)}{f(\gamma(t))} \, dt
= \int_\Gamma \frac{dw}{w} = 2 \pi i \operatorname{N}(\Gamma, 0)
$$
where $\Gamma = f \circ \gamma$ is a closed path, and $\operatorname{N}(\Gamma, 0)$ is the winding number of $\Gamma$ with respect to zero.
The winding number is an integer (see below), and therefore
$$
 \frac{1}{2 \pi i} \int_a^{a+p}\frac{f'(z)}{f(z)} \in \Bbb Z \, .
$$
Why is the winding number an integer? I am pretty sure that
this has been shown before, but I could not find it anymore, therefore
I try to reproduce an elementary proof here.
For $0 \le s \le 1$, define
$$
 F(s) :=  \int_0^s \frac{\Gamma'(t)}{\Gamma(t)} \, dt \, .
$$
Then verify that the derivative of 
$$
  e^{-F(s)}\Gamma(s)
$$
with respect to $s$ is zero. It follows that
$$
 \Gamma(s) = C e^{F(s)} 
$$
for some constant $C$. From $\Gamma(0) = \Gamma(1)$ it follows that
$$
 e^{F(0)} = e^{F(1)}
$$
and therefore
$$
  F(1) = F(0) + 2 \pi i k = 2 \pi i k
$$
for some $k \in \Bbb Z$, which is what we wanted to show.

Re your comments: Here is a direct proof for your special case
$a, a+p \in \Bbb R$, using the same ideas, but without integration 
along paths in $\Bbb C$:
For $x \in \Bbb R$, define 
$$
 F(x) = \int_0^x \frac{f'(t)}{f(t)} \, dt
$$
Then verify that the derivative of
$$
  e^{-F(x)} f(x)
$$
with respect to $x$ is zero. It follows that
$$
  f(x) = C e^{F(x)}
$$
for some constant $C$. From $f(a) = f(a+p)$ it follows that
$$
 e^{F(a)} = e^{F(a+p)}
$$
and therefore 
$$
  F(a+p) = F(a) + 2 \pi i k 
$$
for some $k \in \Bbb Z$. So we have 
$$
\int_a^{a+p} \frac{f'(t)}{f(t)} \, dt = F(a+p) - F(a) = 2 \pi i k \, .
$$
A: Hint
By periodicity, $$\int_a^{a+p}\frac{f'(t)}{f(t)}\mathrm d t\underset{z=f(t)}{=}\int_{f(a)}^{f(a+p)}\frac{1}{z}\mathrm d z\underset{periodicity}{=}\int_{\gamma }\frac{1}{z}\mathrm d z$$
where $\gamma $ is a closed curve. Using cauchy formula, you can prove the claim.
