Evaluate $P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $ where $0 < \alpha <1$

Question

Evaluate
$$P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $$ where $0 < \alpha <1$

Thm Let $P$ and $Q$ be polynomials of degree $m$ and $n$,respectively, where $n \geq m+2$. If $Q(x)\neq 0$. for $Q$ has a zero of order at most 1 at the origin and $f(z)= \frac{z^\alpha P(z)}{Q(z)}$, where $0 < \alpha <1$ then $$P.V, \int^{\infty}_0 \frac{x^ \alpha P(x)}{Q(x)} dx= \frac{2 \pi i}{1- e^{i \alpha 2 \pi }} \sum^{k}_{j=1} Res [f,z_j] $$ where $z_1,z_2 ,\dots , z_k$ are the nonzero poles of $\frac{P}{Q}$

Attempt

Got that $P(x)=1$ where its degree $m=1$ and $q(x)=x(x+1)$ its degree is $n=1$ so it is not the case that $n \geq m+2$ because $2 \geq 1+2$

is the integrand really $$\frac{z^{\alpha}}{x(x+1)}$$? – Dr. Sonnhard Graubner Dec 18 '16 at 19:53 — Dr. Sonnhard Graubner, Dec 18 '16 at 19:53
For $P(x)=1$, $m=0\ne 1$. – Mark Viola Dec 18 '16 at 20:00 — Mark Viola, Dec 18 '16 at 20:00
ok that works then. I need coffe. – Tiger Blood Dec 18 '16 at 20:02 — Tiger Blood, Dec 18 '16 at 20:02

Mark Viola · Answer 1 · 2016-12-18T20:43:30.923

METHODOLOGY $1$: COMPLEX ANALYSIS

Let $f(z)=\frac{z^\alpha}{z(z+1)}$, defined with branch cut along the positive real axis. Let $C$ be the classical "keyhole" contour with the keyhole taken around the branch cut. Then, the residue theorem guarantees that

$$\begin{align} \oint_C f(z)\,dz&=\int_0^\infty \frac{x^\alpha}{x(x+1)}\,dx+\int_\infty^0\frac{x^\alpha\,e^{i2\alpha\pi}}{x(x+1)}\,dx\\\\ &=(1-e^{i2\alpha\pi})\int_0^\infty \frac{x^\alpha}{x(x+1)}\,dx\\\\ &=2\pi i \,\text{Res}\left(\frac{z^\alpha}{z(z+1)}, z=e^{i\pi}\right)\\\\ &=-2\pi i\,e^{i\alpha\pi} \end{align}$$

Hence, we have

$$\int_0^\infty \frac{x^\alpha}{x(x+1)}\,dx=\frac{-2\pi i\,e^{i\alpha\pi}}{1-e^{i2\alpha\pi}}=\frac{\pi}{\sin(\alpha \pi)}$$

METHODOLOGY $2$: REAL ANALYSIS

Alternatively, using the Beta function, we have

$$\begin{align} \int_0^\infty \frac{x^\alpha}{x(x+1)}\,dx&=B\left(\alpha, 1-\alpha\right)\\\\ &=\frac{\Gamma(\alpha)\Gamma(1-\alpha)}{\Gamma(1)}\\\\ &=\frac{\pi}{\sin(\alpha \pi)} \end{align}$$

where we used the relationship between the Gamma and Beta functions along with the Euler's Reflection Formula for the Gamma function.

score 1 · Accepted Answer · answered Dec 18 '16 at 20:05

We assume $0<\alpha<1$. We have

$$ P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx =\frac{\pi}{\sin(\alpha \pi)}. $$

Hint. One may prove that $$ \begin{align} & \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx \\\\&=\int^1_{0} \frac{x^\alpha }{x(x+1)} dx+\int^{\infty}_{1} \frac{x^\alpha }{x(x+1)} dx \\\\&=\int^1_{0} \frac{x^{\alpha-1} }{x+1} dx+\int^0_{1} \frac{x^{\alpha-1} }{1+\frac1x}\cdot \left(- \frac{dx}{x^2}\right) \\\\&=\int^1_{0} \frac{x^{\alpha-1} }{1+x} dx+\int^1_{0} \frac{x^{-\alpha} }{1+x}dx \\\\&=\int^1_{0} \frac{x^{\alpha-1} (1-x)}{1-x^2} dx+\int^1_{0} \frac{x^{-\alpha}(1-x) }{1-x^2}dx \\\\&=\int^1_{0} \frac{x^{\alpha-1} (1-x)}{1-x^2} dx+\int^1_{0} \frac{x^{-\alpha}(1-x)}{1-x^2}dx \\\\&=\frac12\psi\left(\frac{\alpha+1}2\right)-\frac12\psi\left(\frac{\alpha}2\right)+\frac12\psi\left(1-\frac{\alpha}2\right)-\frac12\psi\left(\frac{1-\alpha}2\right) \\\\&=\frac{\pi}{\sin(\alpha \pi)} \end{align} $$ where we have used the classic integral representation of the digamma function $$ \int^1_{0} \frac{1-t^{a-1}}{1-t} dt=\psi(a)+\gamma, \quad a>-1,\tag 1 $$ and the properties $$ \psi(a+1)-\psi(a)=\frac1a,\qquad \psi(a)-\psi(1-a)=-\pi\cot(a\pi). $$

Evaluate $P.V. \int^{\infty}_{0} \frac{x^\alpha }{x(x+1)} dx $ where $0 < \alpha <1$

2 Answers2