finding the series $\sum_{n=1}^\infty \frac{x^n}{n!} \frac{1}{n}$

Question

My goal is to solve this series $$S(x) = \sum_{n=1}^\infty \frac{x^n}{n!} \frac{1}{n}$$

I did took the derivative first w.r.t $x$

$$S'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n!}$$

which I formulated to be $$ S'(x) = \frac{e^x}{x} - \frac{1}{x} $$

Hence, applying the antiderivative

$$ \int \frac{e^x}{x} dx - \ln(x) $$ (I did not put $c$ intentionally)

I have searched for this integral $ \int \frac{e^x}{x} \, dx $ and I found that there's a special function called the exponential integral $$\mathrm{Ei}(x) = \int_{-\infty}^x \frac{e^t}{t} dt$$

I am questioning the feasibility to solve this summation in terms of the $Ei$. I am having a problem regarding the notion of bridging between definite integrals and symbolic antiderivatives ?

EDITED:

There's something just came up to my mind, what if I apply this definite integral on $S'(x)$ $$ \int_1^{x} S'(t) dt = \int_{-\infty}^x \frac{e^t}{t} dt - \int_{-\infty}^1 \frac{e^t}{t} dt - \ln(x) + \ln(1) $$ then, $$ S(x) - S(1) =\mathrm{Ei}(x) - \mathrm{Ei}(1) - \ln(x) $$

which requires finding $S(1) = \sum_{n=1}^{\infty} \frac{1}{n!\,n}$

Answer 1

$S(x)$ is an entire function, it does not have any logarithmic singularity:

$$ S(x)=\sum_{n\geq 1}\frac{x^n}{n\cdot n!}=\int_{0}^{x}\frac{e^t-1}{t}\,dt \tag{1}$$ also since $\frac{e^t-1}{t}$ is an entire function. The RHS of $(1)$ clearly depends on the exponential integral and clearly is not an elementary function.