Explicit formula for the series $ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $ I was wondering if there is an explicit formulation for the series
$$ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k} $$
It is evident that the converges for any $x \in \mathbb{R}$. Any ideas on a formula?
 A: Its derivative is $\sum_{k=1}^{\infty} \frac{x^{k-1}}{k!}$, which is $(e^x-1)/x$.  So your function is the integral of my function, which you might or might not call 'closed form'.
A: You can have the closed form

$$\sum_{k=1}^{\infty}\frac{x^k}{k k!}= -\gamma-\ln(-x)-\Gamma(0, -x), $$

where $\Gamma(s,x)$ is the upper incomplete gamma function. Another possible form is

$$ \sum_{k=1}^\infty \frac{x^k}{k!\cdot k}=-\gamma-\ln  \left( -x \right) -{\it Ei} \left( 1,-x \right),  $$

where 

$$ Ei(a, z) = \int_{1}^{\infty} \frac{e^{-tz}}{t^a}dt,\quad 0 < Re(z),$$

which is known as the exponential integral. The following relation between the exponential integral and the upper incomplete gamma function is useful

$$ Ei(a, z) = z^{a-1}\Gamma(1-a, z). $$

A: Here is my attempt to unravel the mystery of the exponential integral.
\begin{align*}
e^x & = \sum_{k=0}^\infty \frac{x^k}{k!} \\
\frac 1xe^x & = \sum_{k=0}^\infty \frac{x^{k-1}}{k!} \\
\int_{1}^x \frac 1ye^y dy & = \sum_{k=1}^\infty \frac{x^k}{k! \cdot k} + \log x - \sum_{k=1}^\infty \frac{1}{k!k} \\
\int_{1}^x \frac 1ye^y dy - \log x + \sum_{k=1}^\infty \frac{1}{k!k}
& = \sum_{k=1}^\infty \frac{x^k}{k!k}
\end{align*}
This derivation works for $x > 0$, but if $x < 0$, you only need to change the integral from $\int_1^x$ to $\int_{-1}^{x}$ and proceed from there.
