Proof 0f the integral by residue method I have a task of proving the integral $\int_0^\infty \frac1{(x^2 + b^2)^{n+1}}\,dx=\frac{(2n)!\pi}{(n!)^2\,(2b)^{n+1}}$.
By the residue approach I know that I have to find the residue, 
$z=ib$ and $z=-ib$ but by the theorem of semi-circle, I can only take the pole on the upper half of a circle, that's z=ib but I am stuck on how many derivatives to take in order to get this done. 
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Besides the ' Residue Method ', the integration can be performed by a ' Real Method ' as follows:

\begin{align}
\int_{0}^{\infty}{\dd x \over \pars{x^{2} + b^{2}}^{n + 1}} & =
\int_{0}^{\infty}\
\overbrace{{1 \over n!}\int_{0}^{\infty}t^{n}\expo{-\pars{x^{2} + b^{2}}t}
\,\dd t}^{\ds{1 \over \pars{x^{2} + b^{2}}^{n + 1}}}\
\,\dd x =
{1 \over n!}\int_{0}^{\infty}t^{n}\expo{-b^{2}t}\
\overbrace{\int_{0}^{\infty}\expo{-tx^{2}}\,\dd x}^{\ds{{\root{\pi} \over 2}\,t^{-1/2}}}\ \,\dd t
\\[5mm] & =
{\root{\pi}/2 \over n!}\int_{0}^{\infty}t^{n - 1/2}\expo{-b^{2}t}\,\dd t =
{\root{\pi} \over 2\, n!}\bracks{%
{1 \over \pars{b^{2}}^{n + 1/2}}\int_{0}^{\infty}t^{n - 1/2}\expo{-t}\,\dd t}
\\[5mm] & =
{\root{\pi} \over 2\verts{b}^{2n +1}\, n!}\,\Gamma\pars{n + {1 \over 2}}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & =
{\root{\pi} \over 2\verts{b}^{2n +1}\, n!}\,\bracks{%
\root{2\pi}\,2^{1/2 - 2n}\,\Gamma\pars{2n} \over \Gamma\pars{n}}\
\pars{~\Gamma\!-\!Duplication\ Formula~}
\\[5mm] & =
{\pi \over 2^{2n}\verts{b}^{2n +1}\, n!}\,{\pars{2n - 1}! \over \pars{n - 1}!} =
{\pi \over 2^{2n}\verts{b}^{2n +1}\, n!}\,{\pars{2n}!\,/\pars{2n} \over n!\,/n}
\\[5mm] & =
\bbx{\ds{{\pi \over \verts{2b}^{2n +1}}\,{\pars{2n}! \over \pars{n!}^{2}}}}
\end{align}
A: Note that the poles at $z=\pm ib$ are of order $n+1$.  Then, the residues are
$$\begin{align}
2\pi i \text{Res}\left(\frac{1}{(z^2+b^2)^{n+1}}, z=\pm ib\right)&=\frac{2\pi i}{n!}\lim_{z\to \pm ib}\frac{d^n}{dz^n}\left(\frac{(z\mp ib)^{n+1}}{(z^2+b^2)^{n+1}}\right)\\\\
&=\frac{2\pi i}{n!}\lim_{z\to \pm ib}\frac{d^n}{dz^n}\left(\frac{1}{(z\pm ib)^{n+1}}\right)\\\\
&=\frac{2\pi i}{n!}\frac{(-1)^n(n+1)(n+2)\cdots (2n-1)(2n)}{(\pm 2ib)^{2n+1}}\\\\
&=\pm \frac{2\pi (2n)!}{ \,(n!)^2(2b)^{2n+1}}
\end{align}$$
A: You may also notice that complex analysis is not really needed. By assuming $b>0$ and substituting $x=b\tan\theta$ we get:
$$ I = \int_{0}^{+\infty}\frac{dx}{(x^2+b^2)^{n+1}} = \frac{1}{b^{2n+1}}\int_{0}^{\pi/2}\cos(\theta)^{2n}\,d\theta=\frac{1}{2b^{2n+1}}\int_{0}^{\pi}\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{2n}\,d\theta $$
hence:
$$ I = \frac{\pi}{2b^{2n+1}4^n}\binom{2n}{n}=\color{red}{\frac{\pi\,(2n)!}{n!^2(2b)^{2n+1}}}.$$
