4

I tried to compute the following integral: $$ \mbox{For}\quad 0 < a < 1,\qquad \int_{0}^{\infty}\frac{x^{a - 1}}{1 - x}{\rm d}x $$ $$ \mbox{We know that}\quad \int_{0}^{\infty}\frac{x^{a - 1}}{1 - x}{\rm d}x = \int_{0}^{1}\frac{x^{a - 1}}{1 - x}{\rm d}x + \int_{1}^{\infty}\frac{x^{a - 1}}{1 - x}{\rm d}x $$ My strategy is to use the relation between Beta function and Gamma function.

  • But, I can't write $\displaystyle\int_{0}^{1}\frac{x^{a - 1}}{1 - x} {\rm d}x$ as forms of Beta functions: $$ \operatorname{B}\left(c,d\right) = \int_{0}^{1}x^{c - 1}\left(1 - x\right)^{d - 1}{\rm d}x $$ for $\Re\left(c\right)>0,\ \Re\left(d\right) > 0$, since the condition $\Re\left(d\right) > 0$.

Is there any other method to approch this problem ?. Please help me to solve it. Thank you !.

Felix Marin
  • 94,079
wwwwww
  • 162
  • 9

1 Answers1

2

For $0<\Re a<1$, interpreted in the Cauchy principal value sense, the integral \begin{align*} \text{v.p.}\int_0^\infty\frac{x^{a-1}dx}{1-x} &=\lim_{\epsilon\to0^+}\left(\underbrace{\int_0^{1-\epsilon}\frac{x^{a-1}dx}{1-x}}_{x=t}+\underbrace{\int_{1+\epsilon}^\infty\frac{x^{a-1}dx}{1-x}}_{x=t^{-1}}\right) \\&=\lim_{\epsilon\to0^+}\left(\int_0^{1-\epsilon}\frac{t^{a-1}\,dt}{1-t}-\int_0^{1-\epsilon}\frac{t^{-a}\,dt}{1-t}\right)-\underbrace{\lim_{\epsilon\to0^+}\int_{1-\epsilon}^{(1+\epsilon)^{-1}}\frac{t^{-a}\,dt}{1-t}}_{=\ 0} \\&=\int_0^1\frac{t^{a-1}-t^{-a}}{1-t}\,dt=\psi(1-a)-\psi(a)=\color{blue}{\pi\cot\pi a} \end{align*} can indeed be computed the gamma/beta way (the last integral is a particular case of the one here, and the last equality is essentially the reflection formula for the digamma function).

metamorphy
  • 43,591