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 ?

  • 3
    $\begingroup$ 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. $\endgroup$
    – Zacky
    Jan 11, 2019 at 14:44
  • $\begingroup$ 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 ! $\endgroup$ Jan 11, 2019 at 14:50
  • $\begingroup$ Any reason for why a closed form might exist? $\endgroup$ Jan 11, 2019 at 14:56
  • 2
    $\begingroup$ 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 $\endgroup$
    – clathratus
    Jan 11, 2019 at 16:14
  • 1
    $\begingroup$ @HarmonicSun That said, the answer becomes rather nasty as $n$ increases. $\endgroup$ Jan 11, 2019 at 16:15

1 Answer 1


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}

  • $\begingroup$ (+1) Really cool! Is this called contour integration? Would you suggest any resources which I could use to learn this? $\endgroup$
    – clathratus
    Jan 11, 2019 at 23:11
  • $\begingroup$ Great answer, thank you ! $\endgroup$ Jan 11, 2019 at 23:15
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paul Enta
    Jan 11, 2019 at 23:18

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