calculate $\int ^{\infty}_{0}\frac{x^a}{(x+b)^2}dx$ whereas $|a|<1,b>0$

Question

Calculate $\displaystyle \int ^{\infty}_{0}\frac{x^a}{(x+b)^2}\mathrm{d}x$ where $|a|<1$ and $b>0$.

What I thought is taking almost a sphere without a slice on positive real axis.

Answer 1

$x^a$ has a branch line from $0$ to $\infty$. Consider the contour integral from $\infty$ to $0$ below the branch, it will be $-e^{2\pi i a}\int_0^\infty dx \frac{x^a}{(x+b)}$ then continue the contour from $0$ to $\infty$ above the contour, which is simply $\int_0^\infty dx \frac{x^a}{(x+b)}$. The total contour integral is then

$\oint dx \frac{x^a}{(x+b)} = (1-e^{2\pi i a})\int_0^\infty dx \frac{x^a}{(x+b)}$.

The contour can also be closed around the double pole at $-b$, yielding $\oint dx \frac{x^a}{(x+b)} = -2\pi i a e^{\pi i a}b^{a-1}$.

Equating the 2 contour integrals yields $\int_0^\infty dx \frac{x^a}{(x+b)} = \pi\frac{a b^{a-1}}{Sin(\pi a)}$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\int_{0}^{\infty}{x^{a} \over \pars{x + b}^{2}}\,\dd x \,\right\vert_{\ {\large\verts{a}\ <\ 1} \atop {\large\,\,\,\,\,\,\, b\ >\ 0}} & = b^{a - 1}\int_{0}^{\infty}{x^{a} \over \pars{x + 1}^{2}}\,\dd x \\[5mm] & = b^{a}\int_{1}^{\infty}{\pars{x - 1}^{a} \over x^{2}}\,\dd x \\[5mm] & = b^{a - 1}\int_{1}^{0}{\pars{1/x - 1}^{a} \over \pars{1/x}^{2}}\, \pars{-\,{\dd x \over x^{2}}} \\[5mm] & = b^{a - 1}\int_{0}^{1}x^{-a}\pars{1 - x}^{\, a}\,\dd x \\[5mm] & = b^{a - 1}\,{\Gamma\pars{-a + 1}\Gamma\pars{a + 1} \over \Gamma\pars{2}} \\[5mm] & = b^{a - 1}\,\Gamma\pars{-a + 1}\bracks{a\,\Gamma\pars{a}} \\[5mm] & = \bbx{b^{a - 1}\,{\pi a \over \sin\pars{\pi a}}} \end{align}