Evaluar la integral compleja

Question

Evalúa la siguiente integral: $$ \int_{0}^{\infty}{x^{\alpha - 1} \over 1 + x}\,{\rm d}x $$ ¿Cómo empezar?

Answer 1

CONSEJO $$ \int_{0}^{\infty}{x^{\alpha - 1} \over 1 + x}\,{\rm d}x=B(\alpha,1-\alpha)=\Gamma(\alpha)\Gamma(1-\alpha)=\frac{\pi}{\sin \pi\alpha} $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Supondremos que $\ds{0\ <\ a\ <1}$ tal que la integral converge. Partimos de la integral $\ds{\int_{\rm C}{z^{a - 1} \over z + 1}\,\dd z}$ donde $\ds{\rm C}$ es un "contorno de ojo de cerradura" y $$ z^{a - 1}=\verts{z}^{a - 1}\exp\pars{\ic\,{\rm Arg}\pars{z}\pars{a - 1}}\,, \ 0 \ <\ \,{\rm Arg}\pars{z}\ <\ 2\pi\,,\ z\ \not=\ 0 $$

\begin{align} 2\pi\ic\expo{\ic\pi\pars{a - 1}}&=\int_{0}^{\infty}{x^{a - 1} \over x + 1}\,\dd x +\int_{\infty}^{0}{x^{a - 1}\expo{2\pi\ic\pars{a - 1}} \over x + 1}\,\dd x =\pars{1 - \expo{2\pi a\ic}}\int_{0}^{\infty}{x^{a - 1} \over x + 1}\,\dd x \end{align}

$$\color{#66f}{\large\int_{0}^{\infty}{x^{a - 1} \over x + 1}\,\dd x} ={-2\pi\ic\expo{\pi a\ic} \over 1 - \expo{2\pi a\ic}} =\pi\,{2\ic \over -\expo{-\pi a\ic} + \expo{\pi a\ic}} =\pi\,{2\ic \over 2\ic\sin\pars{\pi a}} =\color{#66f}{\large{\pi \over \sin\pars{\pi a}}} $$