Strange symmetry regarding sum $\sum_{n=0}^\infty\frac{n^ne^{-bn}}{\Gamma(n+1)}$ and integral $\int_{0}^\infty\frac{x^xe^{-bx}}{\Gamma(x+1)}dx$ One can show by computation the following for $b>1$
$$\sum_{n=0}^\infty\frac{n^ne^{-b n}}{\Gamma(n+1)}=\frac{1}{1+W_{\color{blue}{0}}(-e^{-b})},\tag{1}$$
(here one assumes that the term with $n=0$ is understood as the limit $\lim_{n\to 0}$ and is equal to $1$) and
$$\int_{0}^\infty\frac{x^xe^{-b x}}{\Gamma(x+1)}dx=\boldsymbol{\color{red}{-}}\frac{1}{1+W_{\color{red}{-1}}(-e^{-b})}.\tag{2}$$
$W_0$ and $W_{-1}$ are different branches of the Lambert W function. One can see that this formulas look similar. I considered them in the hope of obtaining a function for which sum equals integral:
$$
\sum_{n=0}^\infty f(n)=\int_0^\infty f(x) dx.
$$
$(1)$ is the consequence of Lagrange inversion and the integral arises in the probability distribution theory, namely the Kadell-Ressel pdf (see also this MSE post).

Question 1. Can anybody explain the symmetry between $(1)$ and $(2)$ without resorting to direct calculation?
Question 2. Is it possible to alter $(1)$ and $(2)$ to obtain a nice function for which sum equals integral?

If $b=1$ then there is the Knuth series
$$
\sum_{n=1}^\infty\left(\frac{n^ne^{-n}}{\Gamma(n+1)}-\frac1{\sqrt{2\pi n}}\right)=-\frac23-\frac1{\sqrt{2\pi}}\zeta(1/2),\tag{3}
$$
and the "Knuth integral"
$$
\int_0^\infty\left(\frac{x^xe^{-x}}{\Gamma(x+1)}-\frac1{\sqrt{2\pi x}}\right)dx=-\frac13.\tag{4}
$$
Again we see there is a discrepancy.

Question 3. Is it possible to modify the term $\frac1{\sqrt{2\pi x}}$ in $(3)$ and $(4)$ so that the series and the integral agree?

Edit. Of course by mounting some additional terms and parameters one can come up with a formula that technically answers question 2 or 3. What is meant as nice in question 2 might be difficult to formulate explicitly. It is best illustrated by formulas in this MSE post. 
 A: 
Question 2. Is it possible to alter (1) and (2) to obtain a function
  for which sum equals integral?

A simpler form for $z\in[0,\mathrm{e}^{-1})$:
\begin{align}
\sum_{n=0}^\infty
\frac{(z\,n)^n}{\Gamma(n+1)}
&=
\frac1{1+\operatorname{W}_{0}(-z)}
\tag{1}\label{1}
,\\
\int_0^\infty
\frac{(z\,x)^x}{\Gamma(x+1)}\,dx
&=-\frac1{1+\operatorname{W}_{-1}(-z)}
\tag{2}\label{2}
.
\end{align}  
For some $u\in\mathbb{R}$ consider
\begin{align}
\sum_{n=0}^\infty \frac{u}{(n+1)^2}
&=\frac{u\pi^2}6
\tag{3}\label{3}
,\\ 
\int_0^\infty \frac{u}{(x+1)^2}\,dx&=u
\tag{4}\label{4}
.
\end{align}
Let's add \eqref{3} and \eqref{4}
to \eqref{1} and \eqref{2}, respectively:
\begin{align}
 \sum_{n=0}^\infty
 \left(
 \frac{(z\,n)^n}{\Gamma(n+1)}
 +\frac{u}{(n+1)^2}
 \right)
 &=
 \frac1{1+\operatorname{W}_{0}(-z)}
 +\frac{u\pi^2}6
 \tag{5}\label{5}
 ,\\
 \int_0^\infty
 \left(
 \frac{(z\,x)^x}{\Gamma(x+1)} 
 +\frac{u}{(x+1)^2} 
 \right)
 \,dx
 &=-\frac1{1+\operatorname{W}_{-1}(-z)}
 +u
 \tag{6}\label{6}
 .
\end{align}  
From the right hand sides of \eqref{5} and \eqref{6} for any $z\in[0,\mathrm{e}^{-1})$
we have 
\begin{align} 
u&=
-6\frac{2+\operatorname{W_0}(-z)+\operatorname{W_{-1}}(-z)}{(\pi^2-6)(1+\operatorname{W_0}(-z))(1+\operatorname{W_{-1}}(-z))}
\end{align}
such that the pair $(z,u)$ satisfies \eqref{5}=\eqref{6}.
For example, 
\begin{align}
z&=\tfrac12\ln2
,\quad\operatorname{W_0(-z)}=-\ln2,\quad\operatorname{W_{-1}(-z)}=-2\ln2
,\\
&\sum_{n=0}^\infty
\left(
\frac{(n\ln2)^n}{2^n\Gamma(n+1)}
-
\frac{6(2-3\ln2)}{
(\pi^2-6)(1-\ln2)(1-2\ln2)(n+1)^2
}
\right)
\\
=&
\int_{0}^\infty
\left(
\frac{(x\ln2)^x}{2^x\Gamma(x+1)}
-
\frac{6(2-3\ln2)}{
 (\pi^2-6)(1-\ln2)(1-2\ln2)(x+1)^2
}
\right)
\\
=&
\frac{\pi^2(\ln2-1)+6(2\ln2-1)}{
(\pi^2-6)(\ln2-1)(2\ln2-1)
}
\approx 1.549536
.
\end{align}
Edit
Similarly,
\begin{align}
&\sum_{n=0}^\infty
2^{-n}
\left(
\frac{(n\ln2)^n}{\Gamma(n+1)}
+
\frac{\ln2\,(3\ln2-2)}{
 (\ln2-1)(2\ln2-1)^2
}
\right)
\\
=&
\int_{0}^\infty
2^{-x}
\left(
\frac{(x\ln2)^x}{\Gamma(x+1)}
+
\frac{\ln2\,(3\ln2-2)}{
(\ln2-1)(2\ln2-1)^2
}
\right)
\\
=&
\frac{2(\ln2)^2-1}{
(\ln2-1)
(2\ln2-1)^2
}
\approx 0.8537740
.
\end{align}
