How this integral $ \int_0^z\frac{1-e^x}{x} dx$ is connected to the Gamma function and Euler constant? This is my first question in this forum; I hope it is an appropriate question.
The Wolframalpha website tells me that
$$
\int\nolimits_0^z\frac{1-e^x}{x} dx = \log (-z)+\Gamma(0, -z)+\gamma\quad \text{for}\quad \Re(z)<0.
$$
I tried to prove this for myself, but I would appreciate it if you could give 
me some help.
This is not a homework question, and I would appreciate it if someone could point me to a reference or tell me what you expect to happen when $ℜ(z)>0$.
Thanks.
 A: First of all, you misquote WolframAlpha. It give $\int_0^z \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = \gamma + \log(-z) + \Gamma(0, -z)$. Notice $\Gamma(0, -z)$ instead of $\Gamma(0, z)$.
This is done using the fundamental theorem of calculus:
Let
$$
 F(x) = \int \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = \log(x) - \operatorname{Ei}(x) 
$$
Then, $\int_0^z \left( 1- \mathrm{e}^x \right) \frac{\mathrm{d} x}{x} = F(z) - \lim_{x\to 0^+} F(x) = F(z) + \gamma$. The latter limit follows from the Taylor series for the exponential integral.
The connection between $\Gamma(0,-z) = \int_{-z}^\infty \mathrm{e}^{-x} \frac{\mathrm{d} x}{x}$ and $\operatorname{Ei}(z) = \int_{-\infty}^z \mathrm{e}^{x} \frac{\mathrm{d} x}{x}$ is well known.
A: $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...$
$\frac{1-e^{x}}{x}=-1-\frac{x}{2!}-\frac{x^{2}}{3!}...$
$\int_{0}^{z}\frac{1-e^{x}}{x} dx=-z-\frac{z^{2}}{2.2!}-\frac{z^{3}}{3.3!}...$
A: I believe that this works for both positive and negative $z$.
By definition and  change of variables
$$
\Gamma(0,-z)=\int_{-z}^\infty e^{-x}\frac{\mathrm{d}x}{x}=-\int_{-\infty}^z e^x\frac{\mathrm{d}x}{x}\tag{1}
$$
where the principal value is taken where needed. Applying $(1)$ to the integral from $w$ to $z$:
$$
\int_w^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|-\log|w|+\Gamma(0,-z)-\Gamma(0,-w)\tag{2}
$$
According to $(2)$,
$$
\int_0^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|+\Gamma(0,-z)-C\tag{3}
$$
where
$$
C=\lim_{w\to0}(\log|w|+\Gamma(0,|w|))\tag{4}
$$
We can use $\Gamma(0,|w|)$ in $(4)$ since either $\Gamma(0,w)$ or $\Gamma(0,-w)$ is defined by a principal value integral, so we have
$$
\lim_{w\to0}(\Gamma(0,-w)-\Gamma(0,w))=0
$$
Therefore,
$$
\begin{align}
C
&=\lim_{w\to0}(\log|w|+\Gamma(0,|w|))\\
&=\lim_{w\to0}\left(\log|w|+\int_{|w|}^\infty e^{-x}\frac{\mathrm{d}x}{x}\right)\\
&=\lim_{w\to0}\left(\log|w|-\log|w|\;e^{-|w|}+\int_{|w|}^\infty\log(x)e^{-x}\mathrm{d}x\right)\\
&=\int_0^\infty\log(x)e^{-x}\mathrm{d}x\\
&=-\gamma\tag{6}
\end{align}
$$
where $\gamma$ is the Euler-Mascheroni Constant. Combining $(3)$ and $(6)$, we get
$$
\int_0^z\frac{1-e^x}{x}\mathrm{d}x=\log|z|+\Gamma(0,-z)+\gamma\tag{7}
$$
If there is interest, I can append a proof that $\int_0^\infty\log(x)e^{-x}\mathrm{d}x=-\gamma$.
