If $\displaystyle Ei(x) = \int^\infty_x \frac{e^{-u}}{u} du$, using integration by parts we can show that
$\displaystyle Ei(x) = e^{-x} \left( \frac{1}{x} - \frac{1}{x^2} + \frac{2!}{x^3} - \cdots + \frac{(-1)^n (n-1)!}{x^n} \right) + (-1)^n n! \int^\infty_x \frac{e^{-t}}{t^{n+1}} dt$.

Hence obtain an asymptotic expansion of [tex]Ei(x)[tex] as $\displaystyle x \to \infty$. Show that if $\displaystyle x$ is replaced by $\displaystyle z \in \mathbb{C}$, the same asymptotic expansion holds when $\displaystyle |arg z| < \pi$. (Hint choose a suitable contour of integration from $\displaystyle z \to \infty$.)

Is the asymptotic expansion of $\displaystyle Ei(x)$ just
$\displaystyle e^{-x} \left( \frac{1}{x} - \frac{1}{x^2} + \frac{2!}{x^3} - \cdots + \frac{(-1)^n (n-1)!}{x^n} \right)$?

To show that this is also the asymptotic expansion of $\displaystyle Ei(z)$ where $\displaystyle z \in \mathbb{C}$, do I need to prove that the remainder term $\displaystyle (-1)^n n! \int^\infty_z \frac{e^{-t}}{t^{n+1}} dt \to 0$ as $\displaystyle z \to \infty$? What contour should I choose to integrate on?