Calculating $\int_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x$ 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

The reason this all works is because you are only considering the (integrable) singularity in the neighborhood of $x=0$.  Thus, $\alpha \lt 1$ for the integral to converge there.  Infinity is another story.  There the integrand behaves as $x^{-(1+\alpha)}$ ; this leads to the requirement that $\alpha \gt 0$ for convergence.  Thus, $\alpha \in (0,1)$.
To evaluate: evaluate the following integral in the complex plane:
$$\oint_C \frac{dz}{z^{\alpha} (1+z)} $$
where $C$ is a standard keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis.  Thus the contour integral is equal to
$$\int_{\epsilon}^R \frac{dx}{x^{\alpha}(1+x)} + i R^{1-\alpha} \int_0^{2 \pi} d\theta \frac{e^{i (1-\alpha) \theta}}{1+R e^{i \theta}} \\ + e^{-i 2 \pi \alpha} \int_R^{\epsilon}  \frac{dx}{x^{\alpha}(1+x)}+ i \epsilon^{1-\alpha} \int_{2 \pi}^0 d\phi \frac{e^{i (1-\alpha) \phi}}{1+\epsilon e^{i \phi}}$$
As $R \to \infty$, the magnitude of the second integral vanishes.  As $\epsilon \to 0$, the magnitude of the fourth integral vanishes.
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue of the pole at $z=-1$.  Note that here we must use $-1=e^{i \pi}$ due to the choice of contour.  Thus we have
$$\left (1-e^{-i 2 \pi \alpha} \right ) \int_0^{\infty} \frac{dx}{x^{\alpha} (1+x)} = i 2 \pi \, e^{-i \pi \alpha} $$
or

$$\int_0^{\infty} \frac{dx}{x^{\alpha} (1+x)} = \frac{\pi}{\sin{\pi \alpha}}$$

as asserted.
