TITLE: Computing $\int_0^\infty\frac1{(x+1)(x+2)\cdots(x+n)}\mathrm dx $
QUESTION [21 upvotes]: 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
$$\frac1{(x+1)(x+2)\cdots(x+n)}\ ?$$

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

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