calculate $\int ^{\infty}_{0}\frac{x^a}{(x+b)^2}dx$ whereas $|a|<1,b>0$ 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.
 A: $x^a$ 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 $\int_0^\infty dx \frac{x^a}{(x+b)} = \pi\frac{a b^{a-1}}{Sin(\pi a)}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}
