# 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

## 2 Answers


$$\lim_{\epsilon \to 0^{\large +}}\int_{-1}^{1}z^{\epsilon - 1}\dd z = \lim_{\epsilon \to 0^{\large +}}{1 - \expo{\ic\epsilon\pi} \over \epsilon} = \bbx{-\pi\, \ic}$$

• clever choice of cut...+1 – tired May 6 '18 at 14:39
• @AaronHendrickson If you take the indent in the lower plane, you have to change the branch cut definition and, in that case the upper semi-circle cross that branch cut. So, you must integrate with a semi-circle in the lower complex plane such that the branch cut is in upper half complex plane. You have to be careful too with the clockwise and anti-one because it adds an extra sign. – Felix Marin May 6 '18 at 18:19
• @tired Thanks. The OP should state what is the branch cut at the very beginning which clarifies or/and defines the integration meaning. – Felix Marin May 6 '18 at 22:40

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.