Summation or Integral representation ${e^{2}\above 1.5pt \ln(2)}=10.66015459\ldots$ How can I construct a summation or integral representation of $${e^2\above 1.5pt \ln(2)}.$$  Naively I would write $$\Bigg(\sum_{n=0}^{\infty}{2^n \above 1.5pt n!} \Bigg)\Bigg(\sum_{n=1}^\infty {(-1)^{(n+1)} \above 1.5pt n}\Bigg)^{-1}$$ but I suspect we can further reduce that and I am not sure how to get there.  Numerically $${e^{2}\above 1.5pt \ln(2)}=10.66015459\ldots={7.38905609\ldots \above 1.5pt 0.693147181\ldots} $$
 A: There are infinitely many integral representations of $e^2/\ln 2$. Given below is one family of integral representations. 
Let $\int f(x)dx  = F(x)$. The simplest family of solution is when $F(2) = e^2/\ln 2$ and $F(0) = 0$. Once again the simplest family function satisfying these conditions is of the form 
$$
F(x) = \frac{e^x g(x)}{\ln x}
$$
where $g(x)$ is any integrable function such that $g(0) = 0$ and $g(2) = 1$. We can find infinitely many such functions $g(x)$.  For example
$$
g(x) = (x-1)^a, \ \cot\Big(\frac{\pi}{x+2}\Big), \ \frac{\zeta(3)}{\zeta(1+x)}, \ldots
$$
Then 
$$
f(x) = F'(x) = \frac{d}{dx}\bigg(\frac{e^x g(x)}{\ln x}\bigg)
$$
is a function such that $\int_{0}^{2}f(x)dx = e^2/\ln 2$.
For example, taking $g(x) = (x-1)^a, a \ge 1$ gives us the family of solution as
$$
f(x) = F'(x) = \frac{e^x (x-1)^{a-1} \{x(x + a - 1)\ln x- x + 1\} }{x \ln^2 x}.
$$
Thus for all $a \ge 1$, we have
$$
\int_{0}^{2} \frac{e^x (x-1)^{a-1} \{x(x + a - 1)\ln x- x + 1\} }{x \ln^2 x}dx = \frac{e^2}{\ln 2}
$$
A: Note that $1-\log\left(2\right)/e^{2}<1
 $ so $$\sum_{k\geq0}\left(1-\frac{\log\left(2\right)}{e^{2}}\right)^{k}=\frac{1}{1-\left(1-\log\left(2\right)/e^{2}\right)}=\frac{e^{2}}{\log\left(2\right)}$$ for the other part every integral from $0$ to $2$ of the derivative of $$\frac{e^{x}\left(x-1\right)^{\alpha}}{\log\left(x\right)},\alpha\geq1$$ works.
A: Concerning a summation formula, note that we can write
$$
\begin{gathered}
  f(x) = \frac{{x\,e^{\,1 + x} }}
{{\ln \left( {1 + x} \right)}} = e\frac{{x\,e^{\,x} }}
{{\ln \left( {1 + x} \right)}} =  \hfill \\
   = e\sum\limits_{0\, \leqslant \,k} {G\left( k \right)x^{\,k} } \sum\limits_{0\, \leqslant \,j} {\frac{{x^{\,j} }}
{{j!}}}  = e\sum\limits_{0\, \leqslant \,n} {\left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\frac{{G\left( k \right)}}
{{\left( {n - k} \right)!}}} } \right)x^{\,n} }  \hfill \\ 
\end{gathered} 
$$
where the coefficients $G$ are the Gregory's coeffients (with $G(0)=1$ added)
which can be determined in a number of different ways, including this finite sum
involving the (unsigned) Stirling Numbers of 1st kind.
$$
G\left( n \right) = \frac{1}
{{n!}}\sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,n - k} \frac{1}
{{k + 1}}\left[ \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right]} 
$$
Note that for $x=1$ the sum of the $G(n)$ is convergent
$$
\sum\limits_{0\, \leqslant \,n} {G\left( n \right)}  = \frac{1}
{{\ln (2)}}
$$
Starting from the above, various possible sums are expressible.
A: Note really an answer but to big for a comment. Using a BPP substituion for $ln(2)$ we also have $${e^{2}\above 1.5pt ln(2)}=\Bigg(\sum_{n=0}^{\infty}{2^n \above 1.5pt n!}\Bigg)\Bigg({2 \above 1.5pt 3}\sum_{n=0}^{\infty}{1 \above 1.5pt 9^n(2n+1)}\Bigg)^{-1}$$
This is going to be a big stretch for my technical skills but I think we have
$$\Bigg({2 \above 1.5pt 3}\sum_{n=0}^{\infty}{1 \above 1.5pt 9^n(2n+1)}\Bigg)^{-1}=\Bigg({2 \above 1.5pt 3}\Bigg)^{-1}\Bigg(\sum_{n=0}^{\infty}{1 \above 1.5pt 9^n(2n+1)}\Bigg)^{-1}={3 \above 1.5pt 2}\Bigg(\sum_{n=0}^{\infty}{1 \above 1.5pt 9^n(2n+1)}\Bigg)^{-1}$$ 
So at a minimum we can write 
$${e^{2}\above 1.5pt ln(2)}={3 \above 1.5pt 2}\Bigg(\sum_{n=0}^{\infty}{2^n \above 1.5pt n!}\Bigg)\Bigg(\sum_{n=0}^{\infty}{1 \above 1.5pt 9^n(2n+1)}\Bigg)^{-1}$$ Numerically we can write this as
$$10.66015459 ={3 \above 1.5pt 2}e^2(1.039727077083992..)^{-1}=1.5*e^2*0.9617966392597..$$
