 Solution

Choose $\log u$ with $-\pi <\arg u<\pi$. The cut's upper and lower banks have $u=-x$, $x\in (0,1)$, and giveThe small circle about zero vanishes for $\mathrm{Re}z>0$. On the unit circle put $u=e^{i\theta }$. Since $u^2-4u+1=-2u(2-\cos \theta )$, its contribution isUsing $2-\cos \theta =1+2{\sin }^2(\theta /2)$ gives the required decomposition.

The only enclosed pole is $a=2-\sqrt{3}$; the other root $2+\sqrt{3}$ is outside. The residue theorem, including the factor $i$ in front of the contour integral, givesHere the positive real logarithm defines the power, so this expression is an entire function of $z$. The finite-interval integral $G$ is also entire, by differentiation under the integral on compact sets. Therefore $F(z)=\frac{\pi }{\sqrt{3}}(2-\sqrt{3}{)}^{z-1}-G(z)$ is an analytic continuation valid on the whole complex plane. Outside $\mathrm{Re}z>0$ it is this continuation, rather than the unregularized cut integral, that defines $F$.