What's the value of $\sum_{i=1}^\infty \frac{1}{i^2 i!}$? What's the value of $\sum_{i=1}^\infty \frac{1}{i^2 i!}(= S)$?
I try to calculate the value by the following.
$$\frac{e^x - 1}{x} = \sum_{i=1}^\infty \frac{x^{i-1}}{i!}.$$
Taking the integral gives
$$ \int_{0}^x \frac{e^t-1}{t}dt = \sum_{i=1}^\infty \frac{x^{i}}{i i!}. $$
In the same, we gets the following equation
$$ \int_{s=0}^x \frac{1}{s} \int_{t=0}^s \frac{e^t-1}{t}dt ds= \sum_{i=1}^\infty \frac{x^{i}}{i^2 i!}. $$
So we holds
$$S = \int_{s=0}^1 \frac{1}{s} \int_{t=0}^s \frac{e^t-1}{t}dt ds.$$
Does this last integral have an elementary closed form or other expression?
 A: Maybe it's interesting to see how to get the “closed form” in terms of hypergeometric function. Recalling the definition of generalized hypergeometric function $$_{q}F_{p}\left(a_{1},\dots,a_{q};b_{1},\dots,b_{p};z\right)=\sum_{k\geq0}\frac{\left(a_{1}\right)_{k}\cdots\left(a_{q}\right)_{k}}{\left(b_{1}\right)_{k}\cdots\left(b_{p}\right)_{k}}\frac{z^{k}}{k!}
 $$ where $\left(a_{i}\right)_{k}
 $ is the Pochhammer symbol, we note that $\left(2\right)_{k}=\left(k+1\right)!
 $ and $\left(1\right)_{k}=k!$. Hence $$_{3}F_{3}\left(1,1,1;2,2,2;1\right)=\sum_{k\geq0}\frac{\left(k!\right)^{3}}{\left(\left(k+1\right)!\right)^{3}}\frac{1}{k!}=\sum_{k\geq0}\frac{1}{\left(k+1\right)^{3}}\frac{1}{k!}=\sum_{k\geq1}\frac{1}{k^{2}k!}.$$ 
A: By A.S.'s comment, we gets
$$\int_{s=0}^x \frac{1}{s} \int_{t=0}^s \frac{e^t-1}{t}dt ds = \int_{t=0}^x  \frac{e^t-1}{t}\int_{s=t}^x \frac{1}{s}ds dt = \int_{0}^x  \frac{(e^t-1) (\log{x} - \log{t})}{t}dt.$$
So, we holds
$$S = - \int_{0}^1 \frac{(e^t-1) \log{t}}{t}dt = - \int_{- \infty}^0 (e^{e^u}-1) u du.$$
