Cauchy Principal value vs. (usual) improper integral

I am studying the Cauchy principal value. In my complex analysis textbook, I solved the following problem: Evaluate P.V. $\displaystyle \int_0^\infty \frac{x^\alpha}{x(x+1)},$ where $\alpha \in (0,1).$ The Cauchy principal value is $\displaystyle \frac{\pi}{\sin \alpha \pi}.$

I think $\displaystyle \frac{x^\alpha}{x(x+1)}$ is (usual) improper integrable on $(0,\infty)$.

My question is whether we can write $\displaystyle \int_0^\infty \frac{x^\alpha}{x(x+1)}=\frac{\pi}{\sin \alpha \pi}$ or not.

• We usually denote the principal value as $PV\int_{0}^\infty \frac{x^{\alpha}}{x(x+1)}dx=\frac{\pi}{sin(\alpha\pi)}$ But the integral itself is not equal to that expression. – aleden Nov 19 '17 at 16:07
• Cauchy principal value is a way of assigning a result to divergent integrals that is meaningful in some situations. Just like $\sum_{n=1}^\infty n=+\infty$, but $\zeta(-1)=-\frac{1}{12}$ using analytic continuation. – aleden Nov 19 '17 at 16:20