Computing $\int_0^\infty\frac1{(x+1)(x+2)\cdots(x+n)}\mathrm dx $

Question

I would like to compute:

$$\int_0^\infty\frac1{(x+1)(x+2)\cdots(x+n)}\mathrm dx $$ $$ n\geq 2$$

So my question is how can I find the partial fraction expansion of

$$\frac1{(x+1)(x+2)\cdots(x+n)}\ ?$$

Answer 1

HINT: Here's a trick to find partial fraction expansions. Compute

$$\lim_{x\to -k} \frac{(x+k)}{(x+1)(x+2)...(x+n)} \; .$$

This should give you the coefficient of the term $1/(x+k)$ in the expansion.

EDIT: As Américo points out, the partial fraction expansion is

$$\frac{1}{\left( x+1\right) \left( x+2\right) \cdots \left( x+n\right) } =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}\cdot\frac{1}{x+k} \; . $$

The indefinite integral of that expansion is

$$\ln\left( \prod_{k=1}^{n}(x+k)^{\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}} \right) \; .$$

When you fill in the upper bound, you can see that the result must be zero as the leading power in $x$ for the product is $0$ because

$$0 = (1-1)^{n-1} = \sum_{k=0}^{n-1} \frac{(-1)^{k} (n-1)!}{(k)!\left( (n-1)-k\right) !} = (n-1)! \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !} \; .$$

Therefore, we are left with the lower bound

$$-\ln\left( \prod_{k=1}^{n}(k)^{\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}} \right) \; .$$

For $n=2,3$ and $4$ you get resp. $\ln 2$, $\ln(2/\sqrt{3})$ and $\ln(2^5/3^3)/6$.

The lower bound can also be written as

$$\frac{1}{(n-1)!}\sum_{k=0}^{n-1} (-1)^{k-1} {n-1 \choose k} \ln(1+k) \; .$$

Answer 2

If $$\frac{1}{(x+1)(x+2)\dots(x+n)} = \sum_{i=1}^{n} \frac{A_i}{x+i}$$

To compute $A_k$, multiply by $(x+k)$ and set $x = -k$.

In fact, this can be used to show, that for any polynomial $P(x)$ with distinct roots $\alpha_1, \alpha_2, \dots \alpha_n$, that

$$\frac{1}{P(x)} = \sum_{j=1}^{n} \frac{1}{P'(\alpha_j)(x-\alpha_j)}$$

where $P'(x)$ is the derivative of $P(x)$.

Answer 3

Based on my computations in SWP for $2\leq n\leq 8$ I conjecture the following expansion

$$\begin{equation*} \frac{1}{\left( x+1\right) \left( x+2\right) \cdots \left( x+n\right) } =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{(k-1)!\left( n-k\right) !}\cdot\frac{1}{x+k}. \end{equation*}$$

Added. How to prove or disprove? Induction doesn't seem easy.

Added 2. It follows from Aryabhata's answer. See comment below.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &I_{n} \equiv\bbox[10px,#ffd]{\int_{0}^{\infty}{\dd x \over \pars{x + 1}\pars{x + 2}\ldots\pars{x + n}}} = \int_{0}^{\infty}{\dd x \over \pars{x + 1}^{\,\large\overline{n}}} \\[5mm] = &\ \int_{0}^{\infty}{\dd x \over \Gamma\pars{x + 1 + n}/\Gamma\pars{x + 1}} = {1 \over \pars{n - 1}!}\int_{0}^{\infty} {\Gamma\pars{x + 1}\Gamma\pars{n} \over \Gamma\pars{x + n + 1}}\,\dd x \\[5mm] = &\ {1 \over \pars{n - 1}!}\int_{0}^{\infty} \bracks{\int_{0}^{1}t^{x}\pars{1 - t}^{n - 1}\,\dd t}\dd x \\[5mm] = &\ {1 \over \pars{n - 1}!} \int_{0}^{1}\pars{1 - t}^{n - 1}\ \overbrace{\pars{\int_{0}^{\infty}t^{x}\,\dd x}}^{\ds{-\,{1 \over \ln\pars{t}}}}\ \dd t = -\,{1 \over \pars{n - 1}!} \int_{0}^{1}{\pars{1 - t}^{n - 1} \over \ln\pars{t}}\,\dd t \\[5mm] \stackrel{t\ =\ \expo{-z}}{=}\,\,\,& {1 \over \pars{n - 1}!}\int_{0}^{\infty} {\expo{-z}\pars{1 - \expo{-z}}^{n - 1} \over z}\,\dd z \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,& \phantom{-}{1 \over \pars{n - 1}!}\int_{0}^{\infty} \ln\pars{z}\expo{-z}\pars{1 - \expo{-z}}^{n - 1}\,\dd z \\[2mm] & -{1 \over \pars{n - 2}!}\int_{0}^{\infty} \ln\pars{z}\expo{-2z}\pars{1 - \expo{-z}}^{n - 2}\,\dd z \\[5mm] = &\ \phantom{-}{1 \over \pars{n - 1}!}\sum_{k = 0}^{n - 1} {n - 1 \choose k}\pars{-1}^{k} \int_{0}^{\infty}\ln\pars{z}\expo{-\pars{k + 1}z}\,\dd z \\[2mm] & -{1 \over \pars{n - 2}!}\sum_{k = 0}^{n - 2}{n - 2 \choose k} \pars{-1}^{k}\int_{0}^{\infty}\ln\pars{z}\expo{-\pars{k + 2}z} \,\dd z \\[5mm] = &\ -{1 \over \pars{n - 1}!}\sum_{k = 0}^{n - 1} {n - 1 \choose k}\pars{-1}^{k}\,{\ln\pars{k + 1} \over k + 1} \\[2mm] & + {1 \over \pars{n - 2}!}\sum_{k = 0}^{n - 2}{n - 2 \choose k} \pars{-1}^{k}\,{\ln\pars{k + 2} \over k + 2} \\[5mm] = &\ {1 \over \pars{n - 1}!}\sum_{k = 1}^{n} {n - 1 \choose k - 1}\pars{-1}^{k}\,{\ln\pars{k} \over k} + {1 \over \pars{n - 2}!}\sum_{k = 2}^{n}{n - 2 \choose k - 2} \pars{-1}^{k}\,{\ln\pars{k} \over k} \\[5mm] = &\ \bbox[10px,#ffd,border:1px groove navy]{\large\sum_{k = 2}^{n}\mathcal{I}_{nk}\ln\pars{k}}\\ & \end{align} where \begin{equation} \mathcal{I}_{nk} \equiv {\pars{-1}^{k} \over k\pars{n - k}!}\bracks{{1 \over \pars{k - 1}!} + {1 \over \pars{k - 2}!}} = \bbox[10px,#ffd,border:1px groove navy]{\large{\pars{-1}^{k} \over \pars{n - k}!\pars{k - 1}!}}\\ \end{equation}

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$$ \left\{\begin{array}{lcl} \ds{I_{2}} & \ds{=} & \ds{\ln\pars{2}} \\[1mm] \ds{I_{3}} & \ds{=} & \ds{\ln\pars{2} - {1 \over 2}\ln\pars{3}} \\[1mm] \ds{I_{4}} & \ds{=} & \ds{{1 \over 2}\ln\pars{2} - {1 \over 2}\ln\pars{3} + {1 \over 6}\ln\pars{4}} \\[1mm] \ds{I_{5}} & \ds{=} & \ds{{1 \over 6}\ln\pars{2} - {1 \over 4}\ln\pars{3} + {1 \over 6}\ln\pars{4} - {1 \over 24}\ln\pars{5}} \\[1mm] \ds{I_{6}} & \ds{=} & \ds{{1 \over 24}\ln\pars{2} - {1 \over 12}\ln\pars{3} + {1 \over 12}\ln\pars{4} - {1 \over 24}\ln\pars{5} + {1 \over 120}\ln\pars{6}} \\[1mm] \ds{I_{7}} & \ds{=} & \ds{{1 \over 120}\ln\pars{2} - {1 \over 48}\ln\pars{3} + {1 \over 36}\ln\pars{4} - {1 \over 48}\ln\pars{5} + {1 \over 120}\ln\pars{6} - {1 \over 720}\ln\pars{7}} \\ \ds{\vdots} & \ds{=} & \ds{\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots\quad\vdots} \end{array}\right. $$