Does any of you know how to solve
$$\int_0^\infty \frac{u^\alpha}{1+u} du$$
analytically for $\alpha<-1$? E.g., for $\alpha=-\pi$, the solution can be expressed by the Digamma Function according to Wolfram Alpha.
Prior solutions on Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ only cover $0 < \alpha < 1$.
Let $\alpha =-a$, $u=\tan^2 t$, then $$I=2 \int_{0}^{\pi/2} \tan^{1-2a} t~ dt = 2 \int_{0}^{\pi/2} \sin^{1-2a} t ~~ \cos^{2a-1} t~ dt = \Gamma(1-a) \Gamma(a)= \pi \csc \pi a, 0<a<1.$$
We have used Beta-integral and Gamma functions. For $a>1$ the integral diverges. See
https://en.wikipedia.org/wiki/Beta_function