We consider the principal determination of $\displaystyle z^\alpha$ and $\displaystyle \log z$ on $\displaystyle \C\setminus i\R_-$, and we set $\displaystyle f(z)= \frac{z^{\alpha}\log z}{z^2-1}$. We consider the half-circles $\displaystyle \gamma_{a,\lambda}=\{a+\lambda e^{i\theta},0\le \theta\le \pi\}$ in the direct sense, and the closed path $\displaystyle \Gamma$ associated to the following integrals : $\displaystyle I_1= \int_{1+r}^{R}f(x)dx$, et $\displaystyle I_2=\int_{\gamma_{0,R}}f(z)$,
$$\begin{aligned} I_3 & = \int_{[-R,-1-r]}f(z)dz=\int_{R}^{1+r}\frac{e^{\alpha\log(xe^{i\pi})}\log(xe^{i\pi})}{x^2-1} e^{i\pi}dx=e^{i\alpha\pi} \int_{1+r}^{R}\frac{x^{\alpha}(\log x+i\pi)}{x^2-1}dx \\ I_4 & = - \int_{\gamma_{-1,r}}f(z)=- i\pi \frac{i\pi e^{i\alpha\pi}}{-2}=-\frac{\pi^2}{2}e^{i\alpha \pi}\\ I_5 & = \int_{[-1+r,-r]}f(z)dz=e^{i\alpha\pi} \int_{r}^{1-r}\frac{x^{\alpha}(\log x+i\pi)}{x^2-1}dx, \end{aligned}$$ $\displaystyle I_6=-\int_{\gamma_{0,r}}f(z)$, $\displaystyle I_7=\int_{r}^{1-r}f(x)dx,$ et $\displaystyle I_8=-\int_{\gamma_{1,r}}f(z)$.

We have $\displaystyle \int_\Gamma f(z)dz=\sum_{j=1}^7 I_j=0$. Therefore
$$\begin{aligned} 0 & = (1+e^{i\alpha\pi})\left( \int_{1+r}^{R} f(x)dx + \int_{r}^{1-r}f(x)dx \right) + i\pi e^{i\alpha\pi}\left(\int_{r}^{1-r}\frac{x^{\alpha}}{x^2-1}dx+\int_{1+r}^{R}\frac{x^{\alpha}}{x^2-1}dx \right)\\ & + I_2 -\frac{\pi^2}{2}e^{i\alpha \pi} + I_6 + I_8 \\ & \xrightarrow[R\to\infty]{r\to 0} (1+e^{i\alpha\pi}) I + i\pi e^{i\alpha\pi}\int_{0}^{+\infty}\frac{x^{\alpha}}{x^2-1}dx - \frac{\pi^2}{2}e^{i\alpha\pi}. \end{aligned}$$

Hence $\displaystyle (1+e^{-i\alpha\pi})I= \frac{\pi^2}{2}-i\pi \int_{0}^{+\infty}\frac{x^{\alpha}}{x^2-1}dx$. Taking the real part, we obtain: $$I=\frac{\pi^2}{4\cos^2(\alpha\pi/2)}.$$