It is known that $$\delta=-e\left(\gamma+\sum_{k=1}^\infty\frac{(-1)^k}{k\cdot k!}\right)$$Where $\delta$ is the Euler-Gompertz constant and $\gamma$ is the Euler-Mascheroni constant. I want to find expressions of the sum: $$\sum_{k=1}^\infty\frac{(-1)^k}{k\cdot k!}=\varpi$$ So far I found this: $$\varpi=\int_0^\infty e^{1-x}\ln x+e^{-x}\ln(x+1)dx$$ (By substituting $\delta=\int_0^\infty e^{-x}\ln(x+1)dx$ and $\gamma=\int_0^\infty e^{-x}\ln xdx$) And I found out that $$\int e^{e^x}dx=\sum_{k=1}^\infty \frac{e^{xk}}{k\cdot k!}$$So substituting $e^x=-1$ we get that $$\varpi=\int_{-\infty}^{\pi i}e^{e^x}dx$$(not sure if complex numbers are allowed to be on the bounds. Oh well). Please post any other expressions you have found for $\varpi$. A very simple note is that $$\varpi<0$$ Since the first term is negative. Also $\varpi\approx-0.796599599297$
As another note, putting closed forms is optional since it doesn't seem likely. And bounds for $\varpi$ from below are highly appreciated.
Best Answer
Let $$f(x)=\sum_{k=1}^\infty\frac{(-1)^k}{k\cdot k!}x^k. $$ Then $$ f'(x)=\sum_{k=1}^\infty\frac{(-1)^k}{k!}x^{k-1}=\frac1x\sum_{k=1}^\infty\frac{(-1)^k}{k!}x^k=\frac{e^{-x}-1}{x}. $$ So $$ f(1)=\int_0^1\frac{e^{-x}-1}{x}dx=\int_0^1e^{-x}\ln xdx. $$