Issue when evaluate contour integral :$\frac{b^{z-1}}{1-b}$

Question

So, Im evaluating the contour integral $\frac{b^{z-1}}{1-b}$ along the ccw contour encircling the negative half of the real axis. It seems to be analytic in the whole domain, except for at $ b = 0$ and I'm getting a residue of 0 there. I know that the integral is non-zero, though, and i'm wondering where I went wrong. Thanks in advance.

Answer 1

Assume $0 < \Re z <1$, so that your integral converges.

Let $b = -t$, $u = \frac{t}{1+t}$ and notice

$$1 - u = \frac{1}{1+t},\quad t = \frac{u}{1-u}\quad\text{ and }\quad dt = \frac{du}{(1-u)^2}$$

Your integral can be directly evaluated as

$$\begin{align} \int_{-\infty}^{(0+)} \frac{b^{z-1}}{1-b}db = & \lim_{\epsilon\to 0} \left( \int_{-\infty-i\epsilon}^{-i\epsilon} + \int_{i\epsilon}^{-\infty+i\epsilon} \right)\frac{b^{z-1}}{1-b} db \\ = & \int_{\infty}^{0} \frac{ e^{-\pi(z-1)i} t^{z-1}}{1+t} d(-t) + \int_{0}^{\infty} \frac{ e^{\pi(z-1)i} t^{z-1}}{1+t} d(-t)\\ = & (e^{\pi z i} - e^{-\pi z i})\int_0^{\infty} \frac{t^{z-1}}{1+t}dt\\ = & 2i\sin(\pi z) \int_0^1 \left(\frac{u}{1-u}\right)^{z-1} (1-u) \frac{du}{(1-u)^2}\\ = & 2i\sin(\pi z) \int_0^1 u^{z-1}(1-u)^{-z} du\\ = & 2i\sin(\pi z) \frac{\Gamma(z)\Gamma(1-z)}{\Gamma(1)}\\ = & 2i\sin(\pi z) \frac{\pi}{\sin(\pi z)}\\ = & 2\pi i \end{align}$$

Alternatively, when $0 < \Re z < 1$, $|b^{z-1}| \to 0$ as $|b| \to \infty$. You can complete your contour by a clockwise circular loop at $\infty$ and the integral will remain the same. You then deform your new contour to a circle loop around $1$ in clockwise direction. Your integral then becomes

$$\int_{(1-)} \frac{b^{z-1}}{1-b} db = -2\pi i\,\text{Res}\left(\frac{b^{z-1}}{1-b};1\right) = -2\pi i (-1) = 2\pi i$$ Same answer as above.