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

Question

Necesito ayuda para evaluar el integral siguiente.

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

Sé que es necesario utilizar una rama cortado pero no está seguro de cómo empezar.

Cualquier ayuda siempre se agradece.

Editar:

Necesitan ayuda para establecer el principio de Cauchy de la integral impropia si proporciona cualquier dirección.

Answer 1

Suponiendo que $\alpha\in\mathbb{R}$, la integral dada es convergente iff $\alpha\in(-1,0)$.
Con este supuesto tenemos $$ \int{0}^{+\infty}\frac{x^\alpha}{x+1}\,dx = \int{-\infty}^{+\infty}\frac{e^{\alpha z}}{1+e^{-z}}\,dz $ $ y la última integral se puede computar considerando la integral de $f(z)=\frac{e^{\alpha z}}{1+e^{-z}}$ sobre el contorno del rectángulo (orientado en sentido antihorario) tener vértices en $-R,R,R+2\pi i,-R+2\pi i$.
Al computar el residuo en el Polo $z=\pi i$ se desprende que $$ (1-e^{2\pi i\alpha})\int{-\infty}^{+\infty}f(z)\,dz = 2\pi i e^{\pi i\alpha} $ $ que: $$ \int{-\infty}^{+\infty}f(z)\,dz = \color{red}{-\frac{\pi}{\sin(\pi \alpha)}}.$ $

Answer 2

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

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$\ds{\Large\left.a\right)}$ \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}

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$\ds{\Large\left.b\right)}$ \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} Esto es equivalente a la inicial del cambio de variables $\ds{{1 \over x + 1} \mapsto x}$.

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$\ds{\Large\left.c\right)}$ \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}