Exponential integral

Remarks

The substitution $t \mapsto -t$ gives $E_1(x) = -\operatorname{Ei}(-x)$ for $x > 0$. With the Euler–Mascheroni constant $\gamma$, the series $$\operatorname{Ei}(x) = \gamma + \ln \abs{x} + \sum_{k=1}^\infty \frac{x^k}{k \cdot k!}, \qquad E_1(x) = -\gamma - \ln x - \sum_{k=1}^\infty \frac{(-x)^k}{k \cdot k!}$$ hold for $x \ne 0$ and $x > 0$ respectively.

Example: as $x \to \infty$, $E_1(x) \sim \frac{e^{-x}}{x}$, and $E_1(x) = \Gamma(0, x)$ is the upper incomplete gamma function with $s = 0$.

Non-example of a valid argument: $E_1(0)$ is not defined, since $\int_0^1 \frac{e^{-t}}{t}\, dt \ge e^{-1} \int_0^1 \frac{dt}{t} = \infty$.

$E_1$ extends to a holomorphic function on $\C \setminus (-\infty, 0]$.