# $\int_{0}^{\infty} {x^{\alpha} \over \left(x + 1\right)}dx$

I need some help to evaluate the following integral.

$$\int_{0}^{\infty} {x^{\alpha} \over \left(x + 1\right)}dx$$

I know you need to use a branch cut but not sure how to start.

Any help is always appreciated.

Edit:

Need Help to Establish the Cauchy Principle for the improper integral if that provides any direction.

• @Crostul You mean $\alpha \geq 0$ Commented May 3, 2017 at 16:39
• Or $\alpha \leqslant -1$. Commented May 3, 2017 at 16:41

Assuming $\alpha\in\mathbb{R}$, the given integral is convergent iff $\alpha\in(-1,0)$.
With such assumption we have $$\int_{0}^{+\infty}\frac{x^\alpha}{x+1}\,dx = \int_{-\infty}^{+\infty}\frac{e^{\alpha z}}{1+e^{-z}}\,dz$$ and the last integral can be computed by considering the integral of $f(z)=\frac{e^{\alpha z}}{1+e^{-z}}$ over the rectangle contour (counter-clockwise oriented) having vertices at $-R,R,R+2\pi i,-R+2\pi i$.
By computing the residue at the pole $z=\pi i$ it follows that $$(1-e^{2\pi i\alpha})\int_{-\infty}^{+\infty}f(z)\,dz = 2\pi i e^{\pi i\alpha}$$ from which: $$\int_{-\infty}^{+\infty}f(z)\,dz = \color{red}{-\frac{\pi}{\sin(\pi \alpha)}}.$$

• Is there a real-analytic way? Commented May 3, 2017 at 16:50
• @MathematicsStudent1122: of course, through a suitable substitution ($\frac{1}{x+1}=u$), Euler's Beta function and the reflection formula for the $\Gamma$ function. Commented May 3, 2017 at 16:59
• So how would you incorporate the use of Residues Would you set it up $Res( F(z), -1)$ ? Commented May 3, 2017 at 17:05
• @John: the integral over the proposed rectangle contour equals $2\pi i\text{Res}(f(z),z=\pi i)$ by the residue theorem. That leads to the second identity. Commented May 3, 2017 at 17:06
• So, long story short, the substitution $x=e^z$ turns the branch point at $x=-1$ of the original function into a simple pole for my $f(z)$. With such approach the computation of the given integral is greatly simplified. Commented May 3, 2017 at 17:08

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Some alternatives for $\ds{\,\mc{I} \equiv \int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x}$ !!!. $\ds{\qquad \Re\pars{\alpha} \in \pars{-1,0}}$.

$\ds{\Large\left.a\right)}$ \begin{align} 2\pi\ic\pars{1^{\alpha}} & \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, -\int_{-\infty}^{-\epsilon} {\pars{-x}^{\alpha}\expo{-\pi\alpha\ic} \over x - 1}\,\dd x - \int_{\pi}^{-\pi}{\epsilon^{\alpha}\expo{\alpha\theta\,\ic} \over \epsilon\expo{\theta\,\ic} - 1}\,\epsilon\expo{\theta\,\ic}\ic\,\dd\theta - \int_{-\epsilon}^{-\infty} {\pars{-x}^{\alpha}\expo{\pi\alpha\ic} \over x - 1}\,\dd x \\[5mm] & \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{=}\,\,\, \int_{0}^{\infty} {x^{\alpha}\expo{-\pi\alpha\ic} \over x + 1}\,\dd x - \int_{0}^{\infty} {x^{\alpha}\expo{\pi\alpha\ic} \over x + 1}\,\dd x = -2\ic\sin\pars{\pi\alpha}\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x \\[5mm] & \implies \bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}} \end{align}
$\ds{\Large\left.b\right)}$ \begin{align} \mc{I} & = \int_{1}^{\infty}{\pars{x - 1}^{\alpha} \over x}\,\dd x = \int_{1}^{0}{\pars{1/x - 1}^{\alpha} \over 1/x}\,\pars{-\,{1 \over x^{2}}} \,\dd x = \int_{0}^{1}x^{-\alpha - 1}\pars{1 - x}^{\alpha}\,\dd x \\[5mm] & = {\Gamma\pars{-\alpha}\Gamma\pars{\alpha + 1} \over \Gamma\pars{1}} = {\pi \over \sin\pars{-\pi\alpha}} \implies \bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}} \end{align} This one is equivalent to the initial change of variables $\ds{{1 \over x + 1} \mapsto x}$.
$\ds{\Large\left.c\right)}$ \begin{align} \mc{I} & = \int_{0}^{1}{x^{\alpha} \over 1 + x}\,\dd x + \int_{1}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = \int_{0}^{1}{x^{\alpha} \over 1 + x}\,\dd x - \int_{1}^{0}{x^{-\alpha} \over x\pars{1 + x}}\,\dd x \\[5mm] & = \int_{0}^{1}{x^{\alpha} + x^{-\alpha - 1} \over 1 + x}\,\dd x = \int_{0}^{1}{x^{\alpha} + x^{-\alpha - 1} - x^{\alpha + 1} - x^{-\alpha} \over 1 - x^{2}}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1}{x^{\alpha/2 - 1/2} + x^{-\alpha/2 - 1} - x^{\alpha/2} - x^{-\alpha/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\bracks{% -\Psi\pars{{\alpha \over 2} + {1 \over 2}} - \Psi\pars{-\,{\alpha \over 2}} + \Psi\pars{{\alpha \over 2} + 1} + \Psi\pars{-\,{\alpha \over 2} + {1 \over 2}}} \\[5mm] & = {1 \over 2}\braces{% \bracks{\Psi\pars{-\,{\alpha \over 2} + {1 \over 2}} - \Psi\pars{{\alpha \over 2} + {1 \over 2}}} + \bracks{\Psi\pars{{\alpha \over 2} + 1} - \Psi\pars{-\,{\alpha \over 2}}}} \\[5mm] & = {1 \over 2}\bracks{% \pi\cot\pars{\pi\bracks{{\alpha \over 2} + {1 \over 2}}} + \pi\cot\pars{\pi\bracks{-{\alpha \over 2}}}} = -\,{1 \over 2}\,\pi\bracks{% \tan\pars{\pi\bracks{\alpha \over 2}} + \cot\pars{\pi\bracks{\alpha \over 2}}} \\[5mm] & = -\,{1 \over 2}\,\pi\,{1 \over \sin\pars{\pi\alpha/2}\cos\pars{\pi\alpha/2}} \implies \bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}} \end{align}