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QUESTION

By integrating $\ds\frac{z^a}{(z^2+b^2)}$ around a large
semicircle indented at the origin, show that
$$\int_0^\infty\frac{x^adx}{x^2+b^2}=\frac{\pi b^{a-1}}{2\cos
\frac{\pi a}{2}},\ 0<a<1$$

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ANSWER

$\ds I=\int_0^\infty \frac{x^a}{x^2+b^2}\,dx\ \ 0<a<1\\
J=\int_C\frac{z^a}{z^2+b^2}\,dz$

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\begin{eqnarray*}
C & = & \gamma_1+\gamma_2+\gamma_3+\gamma_4\\ \int_{\gamma_1} & =
& I\\ \left|\int_{\gamma_2}\right| & \leq  & \frac{R^a}{r^2-b^2}
\pi R \sim R^{a-1}\to 0\\ & & \textrm{as }R \to \infty \textrm{
for }0<a<1\\ \int_{\gamma_3} & = &
\int_{-\infty}^0\frac{z^a}{z^2+b^2}\,dz\\  & = &
-\int_0^\infty\frac{(-x)^a}{x^2+b^2}(-dx)\\& & (\textrm{taking
}z=-x \textrm{ and swapping the limits of integration})\\ & = &
e^{i\pi a}I\\ \left|\int_{\gamma_4}\right| & \leq & \frac
{r^a}{b^2-r^2}\pi r\sim r^{a+1}\to 0 \textrm{ as } r\to \infty
\textrm{ for } 0<a<1\\ J & = & 2 \pi i \textrm{res}(ib)=2 \pi i
\frac{(be^{i\frac{\pi}{2}})^a}{2ib}\\ & = & \pi b^ae^{\frac{i \pi
a}{2}}\\ I & = &\frac{J}{1+e^{i \pi a}}=\pi b^a \frac{e^{\frac{i
\pi a}{2}}}{1+e^{i \pi a}}\\ & = & \pi b^a\frac{1}{e^{-\frac{i \pi
a}{2}}}+e^\frac{i \pi a}{2}\\ & = & \frac{\pi b^a}{2 \cos
\frac{\pi a}{2}}
\end{eqnarray*}


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