I am looking to write the hypergeometric function $${}_2F_1\left(1,1,2+\epsilon, -\frac{\alpha}{\beta}\right) = \int_0^1\,dt\,\frac{(1-t)^{\epsilon}}{1-tz + i\delta},$$ where $z=-\alpha/\beta$ and $0< \beta < - \alpha$, in terms of its real and imaginary part. The $i\delta$ prescription is to shift the denominator away from the pole at $t=1/z$. I know that $$\frac{1}{1-tz+i\delta} = \text{P.V} \frac{1}{1-tz} -i\pi \delta(1-tz)$$ so to compute the real part I am left with the problem with finding $$\text{P.V}\int_0^1\,dt\,\frac{(1-t)^{\epsilon}}{1-tz}$$ I tried writing this as $$\lim_{\tau \rightarrow 0} \left(\int_0^{1/z-\tau} + \int_{1/z+\tau}^1\right)\frac{(1-t)^{\epsilon}}{1-tz} dt$$ but I am not sure how to progress. I tried using the residue theorem and coming up with a closed contour but the limits do not extend to $\pm \infty$. Any help would be great, thanks!
Edit: I thought maybe I could change variables in the last equation such that the resulting transformation puts the limits on the integration to be $0$ and $1$ and maybe then identify the real part as other (different) hypergeometrics, but I can't see such a transformation yet.