Prove $\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$ for $0\lt a\lt 1$ How could I prove
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\frac{\pi}{2}\tan\frac{a\pi}{2}$$
where $0\lt a\lt 1$?
I only thought of splitting it like this:
$$\int_0^\infty \frac{x^a-1}{x^2-1}\, dx=\int_0^\infty \frac{x^a}{x^2-1}\, dx-\int_0^\infty \frac{dx}{x^2-1},$$
but there's a new unremovable singularity. Partial fraction decomposition would be helpful for natural $a$ but I can't think of anything else here...

Your integral is$$\int_0^1\frac{1-x^a}{1-x^2}dx+\int_0^1\frac{y^{-a}-1}{y^{-2}-1}\frac{dy}{y^2}=\int_0^1\frac{x^{-a}-x^a}{1-x^2}dx.$$For integer $n\ge0$,$$\int_0^1(x^{2n-a}-x^{2n+a})dx=\frac{1}{2n-a+1}-\frac{1}{2n+a+1}=\frac{2a}{(2n+1)^2-a^2}.$$Now finish with a result proven here, with $z=a/2$:$$\sum_{k\ge0}\frac{4a}{(2k+1)^2-a^2}=\pi\tan\frac{a\pi}{2}.$$
