If Ei(x) = \int^\infty_x \frac{e^{-u}}{u} du, using integration by parts we can show that
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 x \to \infty. Show that if x is replaced by z \in \mathbb{C}, the same asymptotic expansion holds when |arg z| < \pi. (Hint choose a suitable contour of integration from z \to \infty.)

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