I have to evaluate $$\int_0^\infty \frac{x^a}{x^2-1}\,dx\qquad (0<a<1)$$ by residual methods.
My approach has been that there are no poles in the upper half-plane but the integrand has a pole at $x = -1$ and at $1$. These have been treated separately and going through the motions, I get a real value (Good, as integral is real!)
However, on checking the value against what Wolfram (and others) give, reveals that this is a divergent integral! Can anyone assist me with an explanation please ?
The integral is divergent at $x=1$, however the limit $$\lim_{\epsilon \to 0^+}\left( \int_0^{1-\epsilon} \frac{x^a}{x^2-1}\,dx + \int_{1+\epsilon}^\infty \frac{x^a}{x^2-1}\,dx \right)$$ does exist, and this is precisely what you are computing with residue methods. One often calls this "integral" the Cauchy principal value. (See here)