sequences and series - Calculate: $sum_{k=1}^infty frac{1}{k}frac{x^k}{k!}$

Find the following sum:

$$\sum_{k=1}^\infty \frac{1}{k}\frac{x^k}{k!}$$

where $x$ is a real number. This is a power series in $x$. In particular, I'm interested in the case $x>0$.

This is very similar to Calculate: $\sum_{k=1}^\infty \frac{1}{k^2}\frac{x^k}{k!}$, the only difference being the $k$ isn't squared in the denominator.

Disclaimer: This is not a homework exercise, I do not know if a closed form solution exists. If it doesn't, exist, then an approximation in terms of well-known functions (not the all-mighty general hypergeometric $_pF_q$, something simpler please) would be desired.

Answer

we have

$$e^{x}-1=\sum _{k=1}^{\infty }{\frac {x^{k}}{k!}}$$
$$\frac{e^{x}-1}{x}=\sum _{k=1}^{\infty }{\frac {x^{k-1}}{k!}}$$

integrate both sides from $x=0$ to $x$

$$\int_{0}^{x}(\frac{e^{x}-1}{x})dx=\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}$$

so

$$\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}=\int_{0}^{x}(\frac{e^{x}-1}{x})dx$$
$$\sum _{k=1}^{\infty }{\frac {x^{k}}{kk!}}=Ei(x)-\log(x)-\gamma$$