Using Contour Integration to Prove $\lim_{\epsilon\to 0^{+}}\int_{-1}^{1}x^{\epsilon-1}\,\mathrm{d}x=-\pi\mathrm{i}$ Let $\epsilon>0$. Then
\begin{equation}
L=\lim_{\epsilon\to0}\int_{-1}^{1}x^{\epsilon-1}\,\mathrm{d}x=-\pi\mathrm{i}.
\end{equation}
This can be proved by direct integration
\begin{equation}
L=\lim_{\epsilon\to0}\frac{x^{\epsilon}}{\epsilon}\bigg|_{x=-1}^{1}=\lim_{\epsilon\to0}\frac{1-(-1)^{\epsilon}}{\epsilon}=-\log(-1)\,\lim_{\epsilon\to0}(-1)^{\epsilon}=-\pi\mathrm{i},
\end{equation}
where we have assumed the principal value of $\log(-1)$. I want to prove this with contour integration.  That said, I tried this by defining the contour $\Gamma$ which extends from $-1$ to $1$ with a semicircular indentation of radius $\delta>0$ into the upper half plane.  This gives the contour integral
\begin{equation}
I
=\lim_{\epsilon\to0}\int_{\Gamma}z^{\epsilon-1}\,\mathrm{d}z
=\lim_{\epsilon\to0}\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}x^{\epsilon-1}\,\mathrm{d}x+\mathrm{i}\int_{\pi}^{0}(\delta e^{\mathrm{i}\varphi})^{\epsilon}\,\mathrm{d}\varphi\right).
\end{equation}
By an argument of uniform convergence, the limit for $\epsilon$ can be brought inside each integral yielding
\begin{equation}
I
=\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}\frac{\mathrm{d}x}{x}+\mathrm{i}\int_{\pi}^{0}\mathrm{d}\varphi\right)=\lim_{\delta\to0}\int_{[-1,1]\setminus[-\delta,\delta]}\frac{\mathrm{d}x}{x}-\mathrm{i}\pi.
\end{equation}
The remaining limit defines a Cauchy principal value integral
\begin{equation}
I
=\mathrm{PV}\int_{-1}^{1}\frac{\mathrm{d}x}{x}-\mathrm{i}\pi=-\pi\mathrm{i}.
\end{equation}
This agrees with the direct integration approach. However, if I change the contour s.t. the semicircular indentation extends into the lower half plane and we follow the same process we get
\begin{equation}
I
=\lim_{\epsilon\to0}\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}x^{\epsilon-1}\,\mathrm{d}x+\mathrm{i}\int_{\pi}^{2\pi}(\delta e^{\mathrm{i}\varphi})^{\epsilon}\,\mathrm{d}\varphi\right)=+\pi\mathrm{i}.
\end{equation}
(1) Is this the result of an error I have made or is the integral path dependent?
(2) If this approach is path dependent, why?
(3) How can I prove this via contour integrals?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Since the integration occurs along $\ds{\pars{-1,1}}$, I'll choose the $\ds{z^{\epsilon - 1}}$-branch cut, with $\ds{\epsilon > 0}$, as
$$
z^{\epsilon - 1} = \verts{z}^{\epsilon - 1}
\expo{\ic\pars{\epsilon - 1}\arg\pars{z}}\,;\qquad z \not= 0\,,\quad -\,{\pi \over 2} < \arg\pars{z} < {3\pi \over 2} 
$$
I'll integrate $\ds{z^{\epsilon - 1}}$ along $\ds{\mc{C}}$ ( a semicircle of unit radius in the upper complex plane which is indented around the origin by an arc of radius $\ds{\delta}$ ). Then, for a given $\ds{\epsilon > 0}$,
\begin{align}
\int_{-1}^{1}z^{\epsilon - 1}\dd z & \equiv
\lim_{\delta \to 0^{+}}\bracks{\int_{-1}^{0}
\pars{-x}^{\epsilon - 1}\expo{\ic\pars{\epsilon - 1}\pi}\dd x +
\int_{\pi}^{0}\bracks{\delta^{\epsilon - 1}\expo{\ic\pars{\epsilon - 1}\theta}}
\delta\expo{\ic\theta}\ic\,\dd\theta +
\int_{0}^{1}x^{\epsilon - 1}\dd x}
\\[5mm] & =
\lim_{\delta \to 0^{+}}\bracks{\overbrace{\oint_{\mc{C}}z^{\epsilon - 1}\dd z}
^{\ds{=\ 0}}\ -\
\underbrace{\int_{0}^{\pi}\bracks{\expo{\ic\pars{\epsilon - 1}\theta}}
\expo{\ic\theta}\ic\,\dd\theta}_{\ds{\mbox{along the circular arc}}}} =
\bbx{1 - \expo{\ic\epsilon\pi} \over \epsilon}
\end{align}

$$
\lim_{\epsilon \to 0^{\large +}}\int_{-1}^{1}z^{\epsilon - 1}\dd z =
\lim_{\epsilon \to 0^{\large +}}{1 - \expo{\ic\epsilon\pi} \over \epsilon} =
\bbx{-\pi\, \ic}
$$
A: When $f(x) = x^{\epsilon - 1}$ is defined on the real axis in terms of the principal value of the logarithm, you cannot continuously extend $f$ from the whole real axis to the lower half-plane. Suppose you start with the positive real axis and go in the clockwise direction. Then the argument of $x$ will be $-\pi$ on the lower bank of the negative real axis. On the other hand, it should be $+\pi$ to make $f$ continuous when starting from the negative real axis and moving in the counterclockwise direction. The values of $e^{i (\epsilon - 1) \arg x}$  will coincide only if $\epsilon$ is an integer.
