# 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 ?

• This integral is so ready for a $x=\frac{1}{t}$ substitution, why not do it? $$\int_0^1 \prod_{k=1}^n \frac{(2k-1)t+1}{2kt +1}dt$$ This looks better. – Zacky Jan 11 at 14:44
• Yes you're right, the $\frac{dx}{x^2}$ was indeed really appealing for this substitution, now I feel really stupid to not having seen it. Thank you for your input ! – Harmonic Sun Jan 11 at 14:50
• Any reason for why a closed form might exist? – Digamma Jan 11 at 14:56
• It might help to note that $$\frac{(2k-1)x+1}{2kx+1}=1-\frac{x}{2kx+1}$$ it sort of looks better to me – clathratus Jan 11 at 16:14
• @HarmonicSun That said, the answer becomes rather nasty as $n$ increases. – Carl Schildkraut Jan 11 at 16:15

To evaluate the integral $$$$I= \int_{1}^{\infty}\frac{1}{x^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+x}\right]\,dx$$$$ we consider the contour integral $$$$J=\int_C \frac{\ln\left( z-1 \right)}{z^2}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\,dz$$$$ $$\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$$. We have then $$$$J=-2i\pi I$$$$ The integrand has a double pole at $$z=0$$ and single poles at $$z=-2p$$, for $$p=1,2,\ldots,n$$. From the residue theorem, $$$$J=2i\pi\left( \operatorname{Res}(z=0)+\sum_{p=1}^n \operatorname{Res}(z=-2p)\right)$$$$ with $$$$\operatorname{Res}(z=0)=\prod_{k=1}^{n}\left[1-\frac{1}{2k}\right]\left.\frac{d\ln(z-1)}{dz}\right|_{z=0}+\ln(-1)\left.\frac{d}{dz}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\right|_{z=0}$$$$ We have \begin{align} \prod_{k=1}^{n}\left[1-\frac{1}{2k}\right]&=\frac{\Gamma(n+1/2)}{\sqrt\pi\Gamma(n+1)}\\ &=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\\ \left.\frac{d}{dz}\prod_{k=1}^{n}\left[1-\frac{1}{2k+z}\right]\right|_{z=0}&=\sum_{k=1}^n\frac{1}{4k^2}\frac{\prod_{p=1}^{n}\left[1-\frac{1}{2p}\right]}{1-\frac{1}{2k}}\\ &=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\sum_{k=1}^n\frac{1}{2k(2k-1)} \end{align} Then $$$$\operatorname{Res}(z=0)=-\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}\left( 1+i\pi\sum_{k=1}^n\frac{1}{2k(2k-1)} \right)$$$$ Now, as the residue at its pole of each of the factors is $$-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$$ is real, we can conclude by taking only the real part of the residues, $$$$I=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}+\sum_{p=1}^n \frac{1}{4p^2}\prod_{k=1\\k\ne p}^{n}\left[1-\frac{1}{2(k-p)}\right]\ln\left( 2p+1 \right)$$$$ which can be written as $$$$I=\frac{2^{1-2n}\Gamma(2n)}{\Gamma(n)\Gamma(n+1)}+\sum_{p=1}^{n-1} \frac{\ln\left( 2p+1 \right)}{p^2}\frac{2^{1-n}\Gamma(2n)\Gamma\left( 2(n-p) \right)}{\Gamma^2(p)\Gamma(n-p)\Gamma(n+1-p)}+\frac{2^{1-2n}\Gamma(2n)}{\Gamma^2(n)}\ln(2n+1)$$$$
• @clathratus Yes this is contour integration. In the case of a rational function to be integrated on the positive real axis, adding a $\ln$ is a classical trick. Many text books of complex analysis describe this kind of methods. I like Henrici "Applied and computational complex analysis", but there are many others. – Paul Enta Jan 11 at 23:18