Evaluating $\int_{1}^{\infty}\frac{1}{x^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+x}\right]\text{d}x$ I'm trying to find a closed form for the following integral, for all $n\in\mathbb{N}^*$ :
$$\int_{1}^{\infty}\frac{1}{x^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+x}\right]\text{d}x$$
As suggested by Zacky, a $x=\frac{1}{t}$ substitution gives :
$$\int_{0}^{1}\prod_{k=1}^{n}\left[\frac{(2k-1)x+1}{2kx+1}\right]\text{d}x$$
The thing is that, for every specific $n\in\mathbb{N}^*$, mathematica is able to compute the integral in terms of logarithms only. In fact, it can always find the antiderivative in terms of logarithms.
What I'm looking for is a general expression for all $n\in\mathbb{N}^*$, if given that it exists.
Now the integrand is an already factorized rational function, so I assumed I could tackle this with some partial fraction decomposition, but I'm not good enough with this method to lead to anything. I'm also not familiar with complex analysis so I haven't tried contour integration, but if you think it can lead to anything, feel free to do so.
Any suggestion ?
 A: To evaluate the integral
\begin{equation}
I= \int_{1}^{\infty}\frac{1}{x^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+x}\right]\,dx
\end{equation} 
we consider the contour integral
\begin{equation}
 J=\int_C \frac{\ln\left( z-1 \right)}{z^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\,dz
\end{equation}
$\ln(z-1)$ is defined with a branch cut from $z=1$ to infinity along the positive real axis.  $C$ is the classical keyhole contour which starts near $z=1$, is then just above the positive real axis, follows the large circle and returns to the starting point, below the real axis and avoids $z=1$. The contribution of the large circle and that of the small circle around $z=1$ vanish. Above the real axis, $z=x+i\varepsilon$ and $\ln\left(z-1 \right)=\ln(x-1)$, and below, $\ln\left(z-1 \right)=\ln(x-1)+2i\pi$. We have then
\begin{equation}
 J=-2i\pi I
\end{equation} 
The integrand has a double pole at $z=0$ and single poles at $z=-2p$, for $p=1,2,\ldots,n$. From the residue theorem,
\begin{equation}
 J=2i\pi\left( \operatorname{Res}(z=0)+\sum_{p=1}^n \operatorname{Res}(z=-2p)\right)
\end{equation} 
with
\begin{equation}
 \operatorname{Res}(z=0)=\prod_{k=1}^{n}\left[1-\frac{1}{2k}\right]\left.\frac{d\ln(z-1)}{dz}\right|_{z=0}+\ln(-1)\left.\frac{d}{dz}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\right|_{z=0}
\end{equation} 
We have
\begin{align}
 \prod_{k=1}^{n}\left[1-\frac{1}{2k}\right]&=\frac{\Gamma(n+1/2)}{\sqrt\pi\Gamma(n+1)}\\
&=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\\
 \left.\frac{d}{dz}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\right|_{z=0}&=\sum_{k=1}^n\frac{1}{4k^2}\frac{\prod_{p=1}^{n}\left[1-\frac{1}{2p}\right]}{1-\frac{1}{2k}}\\
 &=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\sum_{k=1}^n\frac{1}{2k(2k-1)}
\end{align} 
Then
\begin{equation}
 \operatorname{Res}(z=0)=-\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\left( 1+i\pi\sum_{k=1}^n\frac{1}{2k(2k-1)} \right)
\end{equation} 
Now, as the residue at its pole of each of the factors is $-1$
\begin{align}
  \operatorname{Res}(z=-2p)&=-\frac{\ln(2p+1)+i\pi}{4p^2}\prod_{k=1\\k\ne p}^{n}\left[1-\frac{1}{2k-2p}\right]\\
\end{align}
As $I=-J$ and since $I$ is real, we can conclude by taking only the real part of the residues,
\begin{equation}
 I=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}+\sum_{p=1}^n \frac{1}{4p^2}\prod_{k=1\\k\ne p}^{n}\left[1-\frac{1}{2(k-p)}\right]\ln\left( 2p+1 \right)
\end{equation} 
which can be written as
 \begin{equation}
  I=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}+\sum_{p=1}^{n-1} \frac{\ln\left( 2p+1 \right)}{p^2}\frac{2^{1-n}\Gamma(2n)\Gamma\left( 2(n-p) \right)}{\Gamma^2(p)\Gamma(n-p)\Gamma(n+1-p)}+\frac{2^{1-2n}\Gamma(2n)}{\Gamma^2(n)}\ln(2n+1)
 \end{equation}
