Laplace transform II

Laplace transform related to an infinite product

We show the solution of this integral posted on Twitter by @diegorattaggi. The solution turned out to be the logarithm of the golden ratio. We also find a general formulation for this integral wich turned out to be a Laplace transform with a beutiful solution. $$\boxed{\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = \ln\left(\frac{1+\sqrt{5}}{2}\right)}$$ As a corollary we have the Laplace transform: $$\boxed{\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = \ln\left[ \frac{\Gamma\left(\frac{1+s}{5}\right) \Gamma\left(\frac{4+s}{5}\right)}{\Gamma\left(\frac{2+s}{5}\right)\Gamma\left(\frac{3+s}{5}\right)} \right] }$$ Moreover, the integral and the Laplace transform can be further generalized:

If $a_{1}+a_{2}+...+a_{n} = b_{1}+b_{2}+...+b_{n}=c \;$ with $\; a_{i},b_{i} \in \mathbb{N}\setminus\left\{0\right\}$
Hence $$\boxed{ \int_{0}^{\infty} \frac{e^{-a_{1}x}+e^{-a_{2}x}+...+e^{-a_{n}x} -e^{-b_{1}x}-e^{-b_{2}x}-...-e^{-b_{n}x}}{x(1-e^{-cx})}dx = \ln\left(\frac{\Gamma\left(\frac{a_{1}}{c}\right)\Gamma\left(\frac{a_{2}}{c}\right)\cdots\Gamma\left(\frac{a_{n}}{c}\right)}{\Gamma\left(\frac{b_{1}}{c}\right)\Gamma\left(\frac{b_{2}}{c}\right)\cdots\Gamma\left(\frac{b_{n}}{c}\right)}\right)}$$ $$\boxed{ \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-a_{1}x}+e^{-a_{2}x}+...+e^{-a_{n}x} -e^{-b_{1}x}-e^{-b_{2}x}-...-e^{-b_{n}x}}{x(1-e^{-cx})}\right]dx = \ln\left(\frac{\Gamma\left(\frac{a_{1}+s}{c}\right)\Gamma\left(\frac{a_{2}+s}{c}\right)\cdots\Gamma\left(\frac{a_{n}+s}{c}\right)}{\Gamma\left(\frac{b_{1}+s}{c}\right)\Gamma\left(\frac{b_{2}+s}{c}\right)\cdots\Gamma\left(\frac{b_{n}+s}{c}\right)}\right)}$$

Proof
We show the proof of the particular case, the generalization can be shown in a similar manner.

Consider the Laplace transform:
$$\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx$$ Then
$$\frac{d}{ds} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = -\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dx$$ Hence
$$\begin{align*} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dx = & \int_{0}^{\infty} e^{-sx} \sum_{n=0}^{\infty} e^{-5nx} (e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}) dx\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} e^{-x(s+5n)} (e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}) dx\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} \left[ e^{-x(s+5n+1)} -e^{-x(s+5n+2)} - e^{-x(s+5n+3)} + e^{-x(s+5n+4)} \right]dx\\ =& \sum_{n=0}^{\infty}\left[\int_{0}^{\infty} e^{-x(s+5n+1)} dx -\int_{0}^{\infty} e^{-x(s+5n+2)}dx - \int_{0}^{\infty} e^{-x(s+5n+3)}dx + \int_{0}^{\infty} e^{-x(s+5n+4)}dx \right]\\ =& \sum_{n=0}^{\infty}\int_{0}^{\infty} \left[ e^{-x(s+5n-1)} -e^{-x(s+5n-2)} - e^{-x(s+5n-3)} + e^{-x(s+5n-4)} \right]dx\\ =& \sum_{n=0}^{\infty}\left[ \frac{1}{s+5n+1} - \frac{1}{s+5n+2} - \frac{1}{s+5n+3} +\frac{1}{s+5n+4} \right] \end{align*}$$ Then
$$\begin{align*} \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx= &-\int \int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{(1-e^{-5x})}\right] dxds+C\\ =& - \int \sum_{n=0}^{\infty}\left[ \frac{1}{s+5n+1} - \frac{1}{s+5n+2} - \frac{1}{s+5n+3} +\frac{1}{s+5n+4} \right] ds + C\\ =& - \sum_{n=0}^{\infty}\left[ \int \frac{1}{s+5n+1}ds - \int\frac{1}{s+5n+2}ds - \int\frac{1}{s+5n+3}ds +\int\frac{1}{s+5n+4}ds \right] + C\\ =& - \sum_{n=0}^{\infty}\left[ \ln(s+5n+1) - \ln(s+5n+2) -\ln(s+5n+3) +\ln(s+5n+4)\right] + C\\ =& - \sum_{n=0}^{\infty} \ln\left[\frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \end{align*}$$ Therefore
If $\displaystyle s \to 0$ $$\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = - \sum_{n=0}^{\infty} \ln\left[\frac{(5n+1)(5n+4)}{(5n+2)(5n+3)}\right]+C$$ If $\displaystyle s \to \infty$ $$\begin{align*} 0 =& - \lim_{s \to \infty} \sum_{n=0}^{\infty} \ln\left[\frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \\ = & - \sum_{n=0}^{\infty} \ln\left[ \lim_{s \to \infty} \frac{(s+5n+1)(s+5n+4)}{(s+5n+2)(s+5n+3)}\right]+C \\ = & - \sum_{n=0}^{\infty} \ln(1)+C \\ \Longrightarrow C=& 0 \end{align*}$$ Then
$$\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = - \sum_{n=0}^{\infty} \ln\left[\frac{(5n+1)(5n+4)}{(5n+2)(5n+4)}\right] = - \ln\left[ \prod_{n=0}^{\infty} \frac{(5n+1)(5n+4)}{(5n+2)(5n+3)} \right]$$ Recall the following infinite product representation for a quotient of products of Gamma functions (proof in the Appendix):

If $\displaystyle \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}$ then: $$\prod_{k=0}^{\infty}\frac{(a_{1}+k)(a_{2}+k)\cdots(a_{m}+k)}{(b_{1}+k)(b_{2}+k% )\cdots(b_{m}+k)}=\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots% \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)% \cdots\Gamma\left(a_{m}\right)}$$ Hence, we can write:
$$\prod_{n=0}^{\infty} \frac{(5n+1)(5n+4)}{(5n+2)(5n+3)} = \prod_{n=0}^{\infty} \frac{\left(n+\frac{1}{5}\right)\left(n+\frac{4}{5}\right)}{\left(n+\frac{2}{5}\right)\left(n+\frac{3}{5}\right)} = \frac{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}{\Gamma\left(\frac{1}{5}\right) \Gamma\left(\frac{4}{5}\right)} = \frac{2}{1+\sqrt{5}}$$ Therefore
$$\boxed{\int_{0}^{\infty} \frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})} dx = \ln\left(\frac{1+\sqrt{5}}{2}\right)}$$ As a corollary we have the Laplace transform:
$$\boxed{\int_{0}^{\infty} e^{-sx} \left[\frac{e^{-x}-e^{-2x}-e^{-3x}+e^{-4x}}{x(1-e^{-5x})}\right] dx = \ln\left[ \frac{\Gamma\left(\frac{1+s}{5}\right) \Gamma\left(\frac{4+s}{5}\right)}{\Gamma\left(\frac{2+s}{5}\right)\Gamma\left(\frac{3+s}{5}\right)} \right] }$$

Appendix: Euler's infinite product representation for a quotient of Gamma functions

Recall the Euler's definition of the Gamma function $$\Gamma(z) = \lim_{n \to \infty} \frac{(n+1)^z n!}{z(z+1)(z+2)\cdots (z+n)}$$ Supposse that $\displaystyle \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}$ Hence $$\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right) \cdots\Gamma\left(a_{m}\right)} = \lim_{n \to \infty} \prod_{j=1}^{m} \left[(n+1)^{b_{j}-a_{j}} \prod_{k=0}^{n} \frac{a_{j}+k}{b_{j}+k}\right]=\lim_{n \to \infty} (n+1)^{\sum_{j=1}^{m} a_{j} - \sum_{j=1}^{m} b_{j}}\prod_{j=1}^{m} \prod_{k=0}^{n} \frac{a_{j}+k}{b_{j}+k} = \lim_{n \to \infty} \prod_{k=0}^{n} \prod_{j=1}^{m} \frac{a_{j}+k}{b_{j}+k}$$ Therefore: $$\boxed{\prod_{k=0}^{\infty}\frac{(a_{1}+k)(a_{2}+k)\cdots(a_{m}+k)}{(b_{1}+k)(b_{2}+k% )\cdots(b_{m}+k)}=\frac{\Gamma\left(b_{1}\right)\Gamma\left(b_{2}\right)\cdots% \Gamma\left(b_{m}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)% \cdots\Gamma\left(a_{m}\right)} \; \textrm{ if }\; \sum_{j=1}^{m} a_{j} = \sum_{j=1}^{m} b_{j}}$$