# $(-1)^s$ represented in two different ways in the result of integrating along a contour

I'm working through a proof... and at a certain point $(-1)^\frac{1}{2}$ is expressed in two different ways: one as $(e^{i\pi})^\frac{1}{2}$ and another as $(e^{-i\pi})^\frac{1}{2}$.

Here is the context:

$(-t)^\frac{1}{2}$ has a branch cut where t is negative real (so positive x-axis), and the choice of argument is canonical, and it is extended uniquely by continuity throughout the contour. The contour comes in from infinity in the upper half plane towards the origin, winds around the origin, and then leaves towards infinity in the lower half plane.

So the contour is: $H=-H^+ \cup H^o \cup H^-$ where

$H^+:=\{z = w - \epsilon i, w \geq 0\}$ (half line in upper half-plane)

$H^o:=\{z = -e^{i\theta}, \theta \in [-\pi/2,\pi/2]\}$ (semi circle connecting the two half-lines)

$H^-:=\{z = w + \epsilon i, w \geq 0\}$ (half line in lower half-plane)

Now, after calculating the integral, the contribution from the semi-circle vanishes as $\epsilon \rightarrow 0$, and the rest simplifies to the RHS:

$\int_H (-t)^\frac{1}{2}e^{-t}dt = \int_{H^+} (-t)^\frac{1}{2}e^{-t}dt + \int_{H^-} (-t)^\frac{1}{2}e^{-t}dt = (-U+L)\int_0^\infty t^\frac{1}{2}e^{-t}dt$

Where U and L are both determinations of $(-1)^\frac{1}{2}$ based on the choice of argument for $(-t)^\frac{1}{2}$, but as I stated above they are represented in two different ways: one as $(e^{i\pi})^\frac{1}{2}$ (from contour half-Line in lower half-plane) and another as $(e^{-i\pi})^\frac{1}{2}$ (contour half-line in upper half-plane)

I do not understand why U and L are determined the way that they are (in fact I would guess they would be reversed, but it is necessary for the rest of the proof I'm reading to work. Thanks for your help.

• Is "determination" an unusual word for "branch"? – Daniel Fischer Oct 3 '13 at 18:07
• @DanielFischer I've changed the wording to be more clear. – MathStudent Oct 14 '13 at 17:11
• Is there an $i$ missing in some exponent of $e$? – Carsten S Oct 14 '13 at 18:04

That said, if I understand you correctly, your $(\cdot)^{\frac12}$ function is defined by $$(re^{i\phi})^{\frac12}=r^{1/2}e^{i\phi/2},\quad-\pi<\phi<\pi.$$ Hence on the lower half plane, i.e. for $z=re^{i\phi}$ with $-pi<\phi<0$ you have $$(-z)^{\frac12}=(-re^{i\phi})^{\frac12}=(re^{i(\phi+\pi)})^{\frac12}=r^{1/2}e^{i(\phi+\pi)/2} =e^{i\pi/2}r^{1/2}e^{i\phi/2}=e^{i\pi/2}z^{\frac12}=iz^{\frac12}.$$ Note that $-\pi<\phi+\pi<\pi$. On the upper half plane, i.e. for $0<\phi<\pi$ you have $$(-z)^{\frac12}=(-re^{i\phi})^{\frac12}=(re^{i(\phi-\pi)})^{\frac12}=r^{1/2}e^{i(\phi-\pi)/2} =e^{-i\pi/2}r^{1/2}e^{i\phi/2}=e^{-i\pi/2}z^{\frac12}=-iz^{\frac12},$$ because $-\pi<\phi-\pi<\pi$.