asymptotic for the complex exponential integral Ei(s) Let $\operatorname{Ei}(s)$ denote the complex exponential integral function, which (if I am correct) can be defined for $s \in \mathbb{C}\backslash(-\infty,0]$ by
$$\operatorname{Ei}(s) = \gamma+\log s -\operatorname{Ein}(-s)= \gamma+\log s+\sum_{n = 1}^\infty \frac{s^n}{n!n},$$
where $\operatorname{Ein}(s)$ is the entire function $\sum_{n = 1}^\infty \frac{(-1)^{n+1}s^n}{n!n}$.
Is it true that
$$\operatorname{Ei}(s)\sim \frac{e^s}{s} \ (s \to \infty)$$
on $\{s \in \mathbb{C}: |\operatorname{Arg} s|\leq \frac{\pi}{2}-\epsilon\}$ for every $\epsilon > 0$?  More generally, for any $\epsilon > 0$, can one give an asymptotic expansion of $\operatorname{Ei}(s)$ on that set?  Given computations in Mathematica, it seems unlikely that such an asymptotic relation can be extended to the case where $\epsilon = 0$.  (Mathematica's command for the function is $\tt{ExpIntegralEi[s]}$.)
EDIT: I don't know why, but information on the web about the complex function $\operatorname{Ei}(s)$ is very scarce.  But it's an important function used a lot in analytic number theory, and in particular in the Riemann--von Mangoldt explicit formula for $\pi_0(x)$, since one has $\operatorname{li}(s) = \operatorname{Ei}(\log s)$.  Please, somebody, help!
 A: It is known that
$$\tag{1}
E_1 (z) = \int_z^\infty  {\frac{{e^{ - t} }}{t}dt}  \sim \frac{{e^{ - z} }}{z}\left( {1 - \frac{{1!}}{z} + \frac{{2!}}{{z^2 }} - \frac{{3!}}{{z^3 }} +  \cdots } \right)
$$
as $z\to \infty$ in the sector $\left| {\arg z} \right| \le \frac{{3\pi }}{2} - \varepsilon  < \frac{{3\pi }}{2}$ (cf. http://dlmf.nist.gov/6.12.E1). The region of validity is maximal since the rays $\arg z =\pm\frac{{3\pi }}{2}$ are anti-Stokes lines for $E_1 (z)$. Now the function $\operatorname{Ei}(x)$ is defined for positive real $x$ only, and is related to $E_1(z)$ via
$$
\operatorname{Ei}(x) =  - \frac{{E_1 (xe^{\pi i} ) + E_1 (xe^{ - \pi i} )}}{2}.
$$
If you like, you can take $x$ to be complex and by $(1)$, we find the asymptotic series
$$
\operatorname{Ei}(z) \sim \frac{{e^z }}{z}\left( {1 + \frac{{1!}}{z} + \frac{{2!}}{{z^2 }} + \frac{{3!}}{{z^3 }} +  \cdots } \right)
$$
as $z\to \infty$ in the sector $\left| {\arg z} \right| \le \frac{{\pi }}{2} - \varepsilon  < \frac{{\pi }}{2}$, and this region of validity is maximal. To examine what happens on the anti-Stokes line $\arg z =\frac{{\pi }}{2}$ for example, take $z=xe^{\frac{\pi }{2}i}$ with $x>0$ and note that
\begin{align*}
\operatorname{Ei}(xe^{\frac{\pi }{2}i} ) & =   - \frac{{E_1 (xe^{\frac{{3\pi }}{2}i} ) + E_1 (xe^{ - \frac{\pi }{2}i} )}}{2} \\ & =  - \frac{{E_1 (xe^{ - \frac{\pi }{2}i} e^{2\pi i} ) + E_1 (xe^{ - \frac{\pi }{2}i} )}}{2} \\ &=  - E_1 (xe^{ - \frac{\pi }{2}i} ) + \pi i \\ & \sim \pi i + \frac{{e^{ix} }}{{ix}}\left( {1 + \frac{{1!}}{{ix}} + \frac{{2!}}{{(ix)^2 }} + \frac{{3!}}{{(ix)^3 }} +  \cdots } \right)
\end{align*}
as $x\to +\infty$. Here we used http://dlmf.nist.gov/6.4.E2 with $m=1$. Note that the exponential factor is no longer large and hence the previously neglected term $\pi i$ becomes significant.
The complete asymptotic expansion may be written
$$
\operatorname{Ei}(z) \sim \frac{{e^z }}{z}\left( {1 + \frac{{1!}}{z} + \frac{{2!}}{{z^2 }} + \frac{{3!}}{{z^3 }} +  \cdots } \right) + \begin{cases} \quad\, 0, & \text{ if }\; \arg z =0,\\ \;\;\,\pi i, & \text{ if }\; 0<\arg z \leq \frac{\pi}{2}, \\ -\pi i, & \text{ if }\; -\frac{\pi}{2} \leq \arg z <0.\end{cases}
$$
