Summation or Integral representation ${e^{2}\above 1.5pt \ln(2)}=10.66015459\ldots$

Question

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 $${e^{2}\above 1.5pt \ln(2)}=10.66015459\ldots={7.38905609\ldots \above 1.5pt 0.693147181\ldots} $$

Answer 1

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

$$ \int_{0}^{2} \frac{e^x (x-1)^{a-1} \{x(x + a - 1)\ln x- x + 1\} }{x \ln^2 x}dx = \frac{e^2}{\ln 2} $$

Answer 2

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.

Answer 3

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

$$10.66015459 ={3 \above 1.5pt 2}e^2(1.039727077083992..)^{-1}=1.5*e^2*0.9617966392597..$$

Answer 4

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.