# 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 $$L=\lim_{\epsilon\to0}\int_{-1}^{1}x^{\epsilon-1}\,\mathrm{d}x=-\pi\mathrm{i}.$$ This can be proved by direct integration $$L=\lim_{\epsilon\to0}\frac{x^{\epsilon}}{\epsilon}\bigg|_{x=-1}^{1}=\lim_{\epsilon\to0}\frac{1-(-1)^{\epsilon}}{\epsilon}=-\log(-1)\,\lim_{\epsilon\to0}(-1)^{\epsilon}=-\pi\mathrm{i},$$ 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 $$I =\lim_{\epsilon\to0}\int_{\Gamma}z^{\epsilon-1}\,\mathrm{d}z =\lim_{\epsilon\to0}\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}x^{\epsilon-1}\,\mathrm{d}x+\mathrm{i}\int_{\pi}^{0}(\delta e^{\mathrm{i}\varphi})^{\epsilon}\,\mathrm{d}\varphi\right).$$ By an argument of uniform convergence, the limit for $\epsilon$ can be brought inside each integral yielding $$I =\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}\frac{\mathrm{d}x}{x}+\mathrm{i}\int_{\pi}^{0}\mathrm{d}\varphi\right)=\lim_{\delta\to0}\int_{[-1,1]\setminus[-\delta,\delta]}\frac{\mathrm{d}x}{x}-\mathrm{i}\pi.$$ The remaining limit defines a Cauchy principal value integral $$I =\mathrm{PV}\int_{-1}^{1}\frac{\mathrm{d}x}{x}-\mathrm{i}\pi=-\pi\mathrm{i}.$$ 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 $$I =\lim_{\epsilon\to0}\lim_{\delta\to0}\left(\int_{[-1,1]\setminus[-\delta,\delta]}x^{\epsilon-1}\,\mathrm{d}x+\mathrm{i}\int_{\pi}^{2\pi}(\delta e^{\mathrm{i}\varphi})^{\epsilon}\,\mathrm{d}\varphi\right)=+\pi\mathrm{i}.$$

(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?

• integrate over a half circle (which touches the $x$-axis at $\pm1$) and deform the contour to real line for $x>0$ intened around $z=0$ and moving further just a tiny $i\delta\ll i$ above the segement $[-1,0)$. Everything works out only, if you stick to the standard branch of $\log$ – tired May 5 '18 at 18:59
• Could you please clarify? I am having trouble picturing the contour you have described. – Aaron Hendrickson May 5 '18 at 19:18
• this is more or less the same then the approach Felix answer (except the choice of branch cut) – tired May 6 '18 at 14:38

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}
$$\lim_{\epsilon \to 0^{\large +}}\int_{-1}^{1}z^{\epsilon - 1}\dd z = \lim_{\epsilon \to 0^{\large +}}{1 - \expo{\ic\epsilon\pi} \over \epsilon} = \bbx{-\pi\, \ic}$$
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.