TITLE: Prove ${2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}$
QUESTION [6 upvotes]: Prove

$$I={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}\tag1$$

Expand out $(1)$
$$I={2n!\over \pi}\int_{0}^{\infty}{1\over (1+x^2)(2^2+x^2)(3^2+x^2)\cdots(n^2+x^2)}dx\tag2$$
Noticing 
$${1\over (1+x^2)(4+x^2)}={1\over 3}\left({1\over 1+x^2}-{1\over 4+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)}={1\over 2x^2-12x+30}\left({1\over 1+x^2}-{2\over 4+x^2}+{1\over 9+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)(16+x^2)}={1\over -60x^2+300}\left({1\over 1+x^2}-{3\over 4+x^2}+{3\over 9+x^2}-{1\over 16+x^2)}\right)$$
and so on...,
k is a polynomial function of x, we have
(I just realised that if k is a function of x, then what follow from (3) are all wrong!)
$${1\over (1+x^2)(2^2+x^2)\cdots(n^2+x^2)}=k\left({{n-1\choose 0}\over 1+x^2}-{{n-1\choose 1}\over 2^2+x^2}+{{n-1\choose 2}\over 3^2+x^2}-\cdots{{n-1\choose n-1}\over n^2+x^2}\right)\tag3$$
Recall
$$\int_{0}^{\infty}{1\over a^2+x^2}dx={\pi\over 2a}\tag4$$
Sub $(3)$ into $(2)$ and applying $(4)$ hence we have
$$I={2n!k\over \pi}\cdot{\pi\over 2}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag5$$
$$I={n!k}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag6$$
How can we get from $(6)$ to $I={n\over 2n-1}$?
Can anyone produce another method less lengthy than this method above to tackle Integral (1)?
I have saw some authors using the residue theorem to tackle another simple case like the above (1) but I don't know how to apply it.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

The Question:
  $\ds{\quad I \equiv
    {2 \over \pi}\int_{0}^{\infty}\pars{\prod_{k = 1}^{n}
    {k^{2} \over k^{2} + x^{2}}}\,\dd x = {n \over 2n - 1}\,,\qquad n \geq 1}$.

\begin{align}
\color{#f00}{I} & \equiv
{2 \over \pi}\,\pars{n!}^{2}\int_{0}^{\infty}
{\dd x \over \prod_{k = 1}^{n}\pars{x^{2} + k^{2}}}\,\
\stackrel{x^{2}\ \to\ x}{=}\
{\pars{n!}^{2} \over \pi}\,\int_{0}^{\infty}
{x^{-1/2} \over \mathrm{f}\pars{x}}\,\dd x\,,\quad
\left\vert\begin{array}{l}
\mbox{where}
\\
\ds{\mathrm{f}\pars{z} \equiv
\prod_{k = 1}^{n}\pars{z + k^{2}}}
\end{array}\right.
\end{align}

As usual, the integration is performed along a key-hole contour which takes care of the $\ds{z^{-1/2}}$ branch-cut along the 'positive real axis'. Namely,
\begin{align}
\color{#f00}{I} & =
{\pars{n!}^{2} \over \pi}\,\pars{%
\half\,2\pi\ic\,\sum_{k = 1}^{n}
\lim_{z \to -k^{2}}\braces{\bracks{z - \pars{-k^{2}}}{z^{-1/2} \over \mathrm{f}\pars{z}}}}
\\[3mm] & =
\pars{n!}^{2}\,\ic\sum_{k = 1}^{n}
{\verts{-k^{2}}^{-1/2}\pars{\expo{\pi\ic}}^{-1/2} \over
\mathrm{f'}\pars{-k^{2}}} =
\pars{n!}^{2}\sum_{k = 1}^{n}
{1 \over k\,\mathrm{f'}\pars{-k^{2}}}
\end{align}

However,
\begin{align}
k\,\mathrm{f}'\pars{-k^{2}} & =
k\lim_{z\ \to\ -k^{2}}\bracks{%
\mathrm{f}\pars{z}\,\totald{\ln\pars{\mathrm{f}\pars{z}}}{z}} =
k\lim_{z\ \to\ -k^{2}}\bracks{%
\mathrm{f}\pars{z}\,\sum_{\ell = 1}^{n}{1 \over z + \ell^{2}}} =
k\prod_{\ell = 1 \atop {\vphantom{\large A}\ell \not= k}}^{n}
\pars{-k^{2} + \ell^{2}}
\\[3mm] & =
\left\lbrace\begin{array}{lcl}
\ds{\bracks{\prod_{\ell = 2}^{n}\pars{\ell - 1}}
\prod_{\ell = 2}^{n}\pars{\ell + 1}}
& \mbox{if} & \ds{k = 1}
\\[2mm]
\ds{k\bracks{\prod_{\ell = 1}^{k - 1}\pars{\ell - k}}
\bracks{\prod_{\ell = k + 1}^{n}\pars{\ell - k}}
{\prod_{\ell = 1}^{n}\pars{\ell + k} \over 2k}} & \mbox{if} & \ds{k \geq 2}
\end{array}\right.
\\[3mm] & =
\left\lbrace\begin{array}{lcl}
\ds{\pars{n - 1}!\,\pars{3}_{n - 1}} & \mbox{if} & \ds{k = 1}
&&
\\[2mm]
\ds{\half\,\pars{-1}^{k - 1}\
\overbrace{\pars{k - 1}!}^{\ds{\Gamma\pars{k}}}\
\,\pars{1}_{n - k}\,\pars{1 + k}_{n}} & \mbox{if} & \ds{k \geq 2}
\end{array}\right.
\end{align}
where $\pars{a}_{m}$ is a Pochhammer Symbol which satisfies
$\ds{\pars{a}_{m} = {\Gamma\pars{a + m} \over \Gamma\pars{a}}}$. Then,
\begin{align}
k\,\mathrm{f}'\pars{-k^{2}} & =
\left\lbrace\begin{array}{lcl}
\ds{\pars{n - 1}!\,{\Gamma\pars{n + 2} \over \Gamma\pars{3}}}
& \mbox{if} & \ds{k = 1}
\\[2mm]
\ds{%
\half\,\pars{-1}^{k - 1}\Gamma\pars{k}\,
{\Gamma\pars{1 + n - k} \over \Gamma\pars{1}}\,
{\Gamma\pars{1 + k + n} \over \Gamma\pars{1 + k}}}
& \mbox{if} & \ds{k \geq 2}
\end{array}\right.
\\[3mm] & =
\left\lbrace\begin{array}{lcl}
\ds{{n + 1 \over 2n}\,\pars{n!}^{2}}
& \mbox{if} & \ds{k = 1}
\\[2mm]
\ds{%
\half\,{\pars{-1}^{k - 1} \over k}\,\Gamma\pars{1 + n - k}\Gamma\pars{1 + k + n}  }
& \mbox{if} & \ds{k \geq 2}
\end{array}\right.
\end{align}
Indeed, the expression for the case $k = 2$ is $\underline{still\ valid}$ for the value of $k = 1$ such that $\forall\ k \geq 1$
\begin{align}
k\,\mathrm{f}'\pars{-k^{2}} & =
\half\,{\pars{-1}^{k - 1} \over k}\,{\pars{2n}! \over {2n \choose n - k}}
\end{align}
The original integral $\color{#f00}{I}$ is reduced to the finite series
$\pars{~\color{#f00}{\mbox{the final result}}~}$
$$
\begin{array}{|c|}\hline\mbox{}\\
\ds{\
\color{#f00}{I} =
{2 \over \pi}\int_{0}^{\infty}
\pars{\prod_{k = 1}^{n}{k^{2} \over k^{2} + x^{2}}}\,\dd x
=
{2 \over {2n \choose n}}\sum_{k = 1}^{n}\pars{-1}^{k - 1}{2n \choose n - k}k
= 2\,{{2n - 2 \choose n - 1} \over {2n \choose n}} =
\color{#f00}{{n \over 2n - 1}}\ }
\\ \mbox{}\\ \hline
\end{array}
$$

The serie can be evaluated as follows:
  \begin{align}
&{2 \over {2n \choose n}}\sum_{k = 1}^{n}\pars{-1}^{k - 1}{2n \choose n - k}k =
{2 \over {2n \choose n}}\sum_{k = 0}^{\infty}\pars{-1}^{k - 1}k\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{2n} \over z^{n - k + 1}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ {2n \choose n - k}}}
\\[3mm] & =
-\,{2 \over {2n \choose n}}
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{2n} \over z^{n + 1}}\
\overbrace{\sum_{k = 0}^{\infty}k\pars{-z}^{k}}
^{\ds{-\,{z \over \pars{1 + z}^{2}}}}\
\,{\dd z \over 2\pi\ic} =
{2 \over {2n \choose n}}
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{2n - 2} \over z^{n}}
\,{\dd z \over 2\pi\ic}
\\[3mm] & =
{2 \over {2n \choose n}}\,{2n - 2 \choose n - 1} =
2\,{n\pars{n - 1}!n\pars{n - 1}! \over \pars{2n}\pars{2n - 1}\pars{2n - 2}!}
\,{2n - 2 \choose n - 1} = \fbox{$\ds{n \over 2n - 1}$}
\end{align}