Finding a path integral Consider a path
$$ L_1 \ : \quad
\frac{1}{2}+\epsilon \quad \rightarrow \quad
\frac{1}{2}+\epsilon+i(p+\epsilon) \quad \rightarrow \quad
\frac{1}{2}+i(p+\epsilon). $$
where $\epsilon>0$ is arbitrarily small and $p>0$ is fixed.

Question: Find the limit of the path integral
$$ \lim_{\epsilon\to 0^+} \Im\left(\int_{L_1} \frac{d}{ds} \log(s(s-1)) \, ds\right)$$

I tried the following:
First thought: Using fundamental theorem of calculus the path integral becomes
\begin{align*}
\int_{L_1} \frac{d}{ds} \log(s(s-1)) \, ds
&= \log\left[\left(\frac{1}{2}+i(p+\epsilon)\right)\left(-\frac{1}{2}+i(p+\epsilon)\right) \right] \\
&\quad -\log\left[\left(\frac{1}{2}+\epsilon\right)  \left(-\frac{1}{2}+\epsilon\right) \right].
\end{align*}
Hence we have
$$ \int_{L_1} \frac{d}{ds} \log(s(s-1)) \, ds
= \log\left(-(p+\epsilon)^2-\frac{1}{4}\right)-\log\left(\epsilon^2-\frac{1}{4}\right)$$
So we get
$$\lim_{\epsilon\to 0^+} \Im\left(\int_{L_1} \frac{d}{ds} \log(s(s-1)) \, ds\right) = i\pi-i\pi = 0$$
But my Professor says that the answer should be $\pi$. Please help me.
 A: Note that
\begin{align*}
\operatorname{Im} \left( \int_{L_1} \frac{\mathrm{d}}{\mathrm{d}s} \log(s(s-1)) \, \mathrm{d}s \right)
&= \operatorname{Im} \left( \int_{L_1} \left( \frac{1}{s} + \frac{1}{s-1} \right) \, \mathrm{d}s \right) \\
&= \operatorname{Im} \left( \int_{L_1} \frac{1}{s} \, \mathrm{d}s \right) + \operatorname{Im} \left( \int_{L_1} \frac{1}{s-1} \, \mathrm{d}s \right).
\end{align*}
Now we invoke the following well-known fact that
$$ \operatorname{Im}\left( \int_{\gamma} \frac{\mathrm{d}z}{z - a} \right) = \text{[change of argument about $a$ along $\gamma$]} $$
for any piecewise $C^1$ path $\gamma$ not passing through $a$. This gives
\begin{align*}
\operatorname{Im} \left( \int_{L_1} \frac{1}{s} \, \mathrm{d}s \right)
&= \color{blue}{\arctan\left(2(p+\varepsilon)\right)}, \\
\operatorname{Im} \left( \int_{L_1} \frac{1}{s-1} \, \mathrm{d}s \right)
&= \color{red}{-\arctan\left(2(p+\varepsilon)\right)}, \\
\end{align*}
see the picture below as well.

Therefore, we have
$$ \operatorname{Im} \left( \int_{L_1} \frac{\mathrm{d}}{\mathrm{d}s} \log(s(s-1)) \, \mathrm{d}s \right) = 0 $$
and letting $\varepsilon \to 0^+$ still gives $0$.
