My lecturer simply stated the solution to the integral ($\alpha<1, x=O(1)$)

$$I = \int\_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x=\pi\,{\rm cosec}(\pi\alpha)$$

Now the reasoning she gave us that allows us to take this integral even though the integrand doesn't exist at $x=0$ is "we can integrate here since we are only taking leading order". She also said to get this solution she took the first term of the taylor series for $1/(1+x)$. Here is my attempt:

$$I = \int\_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x\approx \int\_0^\infty \frac{1}{x^\alpha}(1)\, {\rm d}x=\left[\frac{x^{1-\alpha}}{1-\alpha}\right]\_0^\infty$$

but this is unbounded so I don't really know how to get the solution. It would be great if someone could help tell me whats wrong or correct her suggestions. Thanks