Solving an improper integral of a product

Question

I am trying to derive a formula for calculating the Euler-Mascheroni constant, which involves the following integral $$\int_{0}^{\infty}\prod_{t=0}^{s}\frac{1}{(x+t+1)}dx$$ For $s\ge 1$.

My first thought was to use partial fraction decomposition with the Heaviside method. $$\prod_{t=0}^{s}\frac{1}{(x+t+1)}=\sum_{t=0}^{s}\frac{C_t}{x+t+1}=\sum_{t=0}^{s}\left(\frac{1}{x+t+1}\prod_{w=0\\w\neq t}^{s}\frac{1}{(w-t)}\right)$$ Then, the integral becomes $$\int_{0}^{\infty}\sum_{t=0}^{s}\left(\frac{1}{x+t+1}\prod_{w=0\\w\neq t}^{s}\frac{1}{(w-t)}\right)dx$$ However, I can't switch the integral and summation operations because then the integral diverges.

Answer 1

For every integer $n\ge 2$ define the function $f_n$ on $[0,\infty)$ by $$ f_n(x)=\prod_{i=1}^{n}\frac{1}{x+i}=\sum_{i=1}^{n}\frac{c_i}{x+i} $$ where \begin{eqnarray} c_k&=&\left(\prod_{i=1}^{k-1}\frac{1}{-k+i}\right)\left(\prod_{i=k+1}^{n}\frac{1}{-k+i}\right)\cr &=&\frac{(-1)^{k-1}}{[1\cdot2\cdots(k-1)]\cdot[1\cdot2\cdots(n-k)]}\cr &=&\frac{(-1)^{k-1}k}{k!\cdot (n-k)!}\cr &=&\frac{(-1)^{k-1}k}{n!}{n \choose k} \end{eqnarray} Define $$ a_k=(-1)^{k-1}k{n\choose k} $$ and $$ \sigma_k=\sum_{i=1}^k a_i $$ Since $$\tag{1} (1-x)^n=\sum_{k=0}^n{n\choose k}(-x)^k $$ Taking the derivative of (1) we have $$\tag{2} -n(1-x)^{n-1}=-\sum_{k=1}^nk{n\choose k}(-x)^{k-1} $$ Setting $x=1$ in (2), we get
$$ 0=-\sum_{k=1}^n(-1)^{k-1}k{n\choose k}=-\sigma_n $$ i.e. $\sigma_n=0$.

We have \begin{eqnarray} n!\cdot f_n(x)&=&\sum_{i=1}^n\frac{a_i}{x+i}\cr &=& \sigma_1\left(\frac{1}{x+1}-\frac{1}{x+2}\right)+\frac{\sigma_2}{x+2}+\sum_{i=3}^n\frac{a_i}{x+i}\cr &\vdots&\cr &=&\sum_{i=1}^{n-1}\sigma_i\left(\frac{1}{x+i}-\frac{1}{x+i+1}\right)+\frac{\sigma_n}{x+n}\cr &=&\sum_{i=1}^{n-1}\sigma_i\left(\frac{1}{x+i}-\frac{1}{x+i+1}\right) \end{eqnarray}

It follows that \begin{eqnarray} n!\int_0^{\infty}f_n(x)dx&=&\sum_{k=1}^{n-1}\sigma_k\ln\left(\frac{x+k}{x+k+1}\right)\Big|_0^{\infty}\cr &=&-\sum_{k=1}^{n-1}\sigma_k\ln\left(\frac{k}{k+1}\right)\cr &=&\sum_{k=1}^{n-1}\sigma_k\ln(k+1)-\sum_{k=1}^{n-1}\sigma_k\ln(k)\cr &=&\sum_{k=2}^n\sigma_{k-1}\ln(k)-\sum_{k=2}^{n-1}\sigma_k\ln(k)\cr &=&\sigma_{n-1}\ln(n)-\sum_{k=2}^{n-1}[\sigma_k-\sigma_{k-1}]\ln(k)\cr &=&-a_n\ln(n)-\sum_{k=2}^{n-1}a_k\ln(k)\cr &=&-\sum_{k=2}^na_k\ln(k). \end{eqnarray} Hence $$ \int_0^{\infty}\prod_{k=1}^n\frac{1}{x+k}dx=\sum_{k=2}^{n}\frac{(-1)^kk{n \choose k}\ln(k)}{n!} $$

Answer 2

Note that, $$\prod_{t=0}^{s}\frac{1}{x+t+1}=\prod_{t=x+1}^{x+s+1}\frac{1}{t}$$ Which further simplifies to, $$\frac{\Gamma(x)}{\Gamma(x+s+2)}$$ So we have, $$\int_{0}^{\infty}\frac{\Gamma(x)}{\Gamma(x+s+2)}dx$$ $$\frac{1}{\Gamma(s+2)}\int_{0}^{\infty}B(x,s+2)dx$$ $$\frac{1}{\Gamma(s+2)}\int_{0}^{\infty}\int_{0}^{1}t^{x-1}(1-t)^{s+1}dt dx$$ $$\frac{1}{\Gamma(s+2)}\int_{0}^{1}\frac{(1-t)^{s}}{t}\int_{0}^{\infty}t^{x}dxdt$$ The Integral $\int_{0}^{\infty}t^{x}\mathrm{dx}$ is easy to evaluate as $0≤t≤1$.

I hope you can proceed from here as per your problem.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$ \begin{align} & \color{#44f}{\left.\int_{0}^{\infty}\prod_{t = 0}^{s}{1 \over x + t + 1}\,\dd x \,\right\vert_{\, s\ \geq\ 1}} = \int_{0}^{\infty}\pars{\sum_{t = 0}^{s}{a_{t} \over x + t + 1}}\dd x \end{align} \begin{align} \mbox{where}\quad a_{t} & \equiv \lim_{x \to -t - 1}\,\, \bracks{\pars{x + t + 1}{\Gamma\pars{x + 1} \over \Gamma\pars{x + s + 2}}} \\[5mm] & = -\pi\lim_{x \to -t - 1}\,\, \bracks{{x + t + 1 \over \sin\pars{\pi x}}{1 \over \Gamma\pars{-x}\Gamma\pars{x + s + 2}}} \\[5mm] & = {\pars{-1}^{t} \over t!\,\,\pars{s - t}!}.\quad \mbox{Note that}\ \left.\sum_{t = 0}^{s}a_{t}\right\vert_{s\ \geq\ 1} = 0 \end{align} Then, \begin{align} & \color{#44f}{\left.\int_{0}^{\infty}\prod_{t = 0}^{s}{1 \over x + t + 1}\,\dd x \,\right\vert_{\, s\ \geq\ 1}} = \lim_{\Lambda\, \to\, \infty}\ \sum_{t = 0}^{s} a_{t}\ln\pars{\Lambda + t + 1 \over t + 1} \\[5mm] = & \ \lim_{\Lambda\, \to\, \infty}\ \sum_{t = 0}^{s} a_{t}\bracks{\ln\pars{\Lambda} + % \ln\pars{1 + {t + 1 \over \Lambda}} -\ln\pars{t + 1}} \\[5mm] = & \ \bbx{\color{#44f}{-\sum_{t = 0}^{s}\pars{-1}^{t} {\ln\pars{t + 1} \over t!\,\,\pars{s - t}!}}} \\ & \end{align}