Evaluate this integral $\int ^\infty _0 \frac{z^{a-1}}{z^2+b}dz$ Evaluate this integral $$\int ^\infty _0 \frac{z^{a-1}}{z^2+b}dz \;\;\;\; 0<a<2,b>0$$
My attempt:
here the zeros of $z^2+b=0 \Rightarrow z=\pm bi$
and $z=bi$ is the upper Half plan  and it is only pole of order one so,
$Res_{z=bi}f(z)=\lim _{z\to bi} \{(z-ib)\frac{z^{a-1}}{z^2+b}\}\\
=\lim _{z\to bi} \{(z-ib)\frac{z^{a-1}}{z+ib}\}$  
is i a m going right way..need help thank you..........
 A: Hint:
$$
\int_0^\infty\frac{z^{a-1}}{z^2+b}\,\mathrm{d}z
$$
Substitute $z=\sqrt{bx}$, then look at this answer.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\left.\int_{0}^{\infty}{z^{a - 1} \over z^{2} + b}\,\dd z
\,\right\vert_{\large\ {0\ <\ \Re\pars{a}\ <\ 2 \atop b\ >\ 0}} &
\,\,\,\stackrel{z\ =\ b^{\large 1/2}\,t}{=}\,\,\,
\int_{0}^{\infty}{b^{\pars{a - 1}/2}\,t^{a - 1} \over bt^{2} + b}\,b^{1/2}
\,\dd t =
b^{a/2 - 1}\int_{0}^{\infty}{t^{a - 1} \over t^{2} + 1}\,\dd t
\\[5mm] &
\stackrel{t^{2}\ \mapsto\ t}{=}\,\,\,
{1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}{t^{a/2 - 1} \over t + 1}\,\,\dd t
\\[5mm] & =
{1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}t^{a/2 - 1}\int_{0}^{\infty}\expo{-\pars{t + 1}x}\,\dd x\,\dd t
\\[5mm] & =
{1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}\expo{-x}\
\underbrace{\int_{0}^{\infty}t^{a/2 - 1}\expo{-xt}\,\dd t\,\dd x}
_{\ds{\Gamma\pars{a/2} \over x^{a/2}}}
\\[5mm] & =
{1 \over 2}\,b^{a/2 - 1}\,\Gamma\pars{a \over 2}
\int_{0}^{\infty}x^{-a/2}\expo{-x}\,\dd x
\\[5mm] & =
{1 \over 2}\,b^{a/2 - 1}\,\Gamma\pars{a \over 2}\Gamma\pars{-\,{a \over 2} + 1} =
\bbx{{1 \over 2}\,b^{a/2 - 1}\,{\pi \over \sin\pars{\pi a/2}}}
\end{align}
