How to prove the closed form of the integral $\int \frac {dx}{\prod_{r=0}^n (x+r)}$

Solution 1:

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int{\dd x \over \prod_{r = 0}^{n}\pars{x + r}} & = \int{\dd x \over \Gamma\pars{x + n + 1}/\Gamma\pars{x}} = {1 \over n!} \int{\Gamma\pars{x}\Gamma\pars{n + 1}\over \Gamma\pars{x + n + 1}}\,\dd x \\[5mm] & = {1 \over n!}\int\int_{0}^{1}t^{x - 1}\pars{1 - t}^{n}\,\dd t\,\dd x = {1 \over n!}\int_{0}^{1}\pars{1 - t}^{n}\int t^{x - 1}\,\dd x \\[5mm] & = {1 \over n!}\int_{0}^{1}\pars{1 - t}^{n} \bracks{{t^{x - 1} \over \ln\pars{t}} + \,\mrm{A}\pars{t}}\,\dd x \\[5mm] & = {1 \over n!}\int_{0}^{1}{t^{x - 1}\pars{1 - t}^{n} \over \ln\pars{t}}\,\dd t + {1 \over n!}\int_{0}^{1}{t^{x - 1}\,\mrm{A}\pars{t} \over \ln\pars{t}}\,\dd t \end{align}