Integral $\int_0^\infty \frac{u^\alpha}{1+u} du$ for $\alpha<-1$

Question

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 $0 < \alpha < 1$.

Are you sure about the result for $\alpha = \pi$? The integral should not converge in that case. — Cardboard Box, Feb 15 '20 at 14:31
I think the integral still diverges at $\alpha=-\pi$. As a Beta integral of the second kind (5th property here), it surely converges for $\alpha\in(-1,0)$. — StubbornAtom, Feb 15 '20 at 15:18

Z Ahmed · Accepted Answer · 2020-02-15T15:42:54.510

2

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

edited Feb 15 '20 at 15:42

answered Feb 15 '20 at 14:45

Z Ahmed

43,235

How do you go from $\infty$ to $\pi/2$? – Jenny Reininger Feb 15 '20 at 14:58
@JennyReininger what is $u$ when $t = \infty$? – an4s Feb 15 '20 at 15:26
When $u=0, t=0$; when $u=\infty, t=\pi/2$ as $u=\tan^2 t$. – Z Ahmed Feb 15 '20 at 15:45

Integral $\int_0^\infty \frac{u^\alpha}{1+u} du$ for $\alpha<-1$

1 Answers1