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\newcommand\ds{\displaystyle}

QUESTION

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

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ANSWER

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

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


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