Question about the asymptotic expansion of the Exponential Integral $\int_z^\infty \frac{e^{-x}}{x}dx$: using Taylor expansion on the infinite domain

Question

I am reading the argument on finding the first three terms of the asymptotic series of the Exponential integral $E_1(z)$ as $z\to \infty$, but I don't understand a step here.

The Exponential integral is given by $E_1(z)=\int_z^\infty \frac{e^{-x}}{x}dx$.

The proof proceeds as follows: Let $x=z+t, dx=dt.$

$$E_1(z)=\int_z^\infty \frac{e^{-x}}{x}dx=\int_0^\infty \frac{e^{-z-t}}{z+t}dt = \frac{e^{-z}}{z}\int_0^\infty \frac{e^{-t}}{1+t/z}dt \\ = \frac{e^{-z}}{z}\int_0^\infty e^{-t}\big(1-\frac{t}{z}+\frac{t^2}{z^2}+\cdots\big)dt.$$

Note, this expansion is valid provided $|t|<|z|,$ but for $|t|>|z|,$ the error is exponentially small due to $e^{-t}$ term, so OK to use for all $t$. Also, we know that $$I_n=\int_0^\infty t^n e^{-t}dt=[t^{-n}e^{-t}]_0^\infty + \int_0^\infty nt^{n-1}e^{-t}dt=n!.$$

So we have $$E_1(z)\sim \frac{e^{-z}}{z}\big(1-\frac{1}{z}+\frac{2!}{z^2}+\cdots\big).$$

I can't grasp why it is OK to use replace the $\frac{1}{1+t/z}$ in the integrand by its Taylor expansion. As it says, for $|t|<|z|$ we can do this, but I can't see why we can replace it in the entire domain. By saying the error is exponentially small, does it mean that $\int_z^\infty e^{-t}(1-t/z+t^2/z^2+\cdots) - e^{-t}(1+t/z)^{-1}dt = O(e^{-z})$? But why is this? I can't see why the $(1-t/z+t^2/z^2+\cdots)$ can be used as an integrand and integrated term by term when it does not uniformly converge in $t>z$. And why does the exponentially small error that the terms we get by doing the integral $\int_0^\infty e^{-t}(1-t/z+t^2/z^2+\cdots)$ will be the asymptotic series and we can ignore the exponentially small errors?

Answer 1

You can use the following formula always:

$$\frac1{1 + r} = \sum_{n=0}^N (-1)^n r^n + \frac{(-1)^{N+1} r^{N+1}}{1+r}$$

It is not hard to prove it. Just compute the sum in the middle as a finite geometric sum. Now use the formula for $r = t/z$,

\begin{align} \int_z^{\infty} \frac{e^{-x}}{x}\mathrm d x &= \frac{e^{-z}}{z}\int_0^{\infty} \frac{1}{1+t/z} e^{-t}\mathrm d t\\ &= \frac{e^{-z}}{z} \left(\sum_{n=0}^{N} \frac{(-1)^n}{z^{n}}\int_{0}^{\infty}t^ne^{-t}\mathrm d t + \frac{(-1)^{N+1}}{z^{N+1}}\underbrace{\int_0^\infty \frac{t^{N+1}}{1+t/z} e^{-t}\mathrm dt}_{=\mathcal O(1)}\right)\\ &= \sum_{n=0}^{N} (-1)^n n!\frac{e^{-z}}{z^{n+1}} + \underset{z\to\infty}{\mathcal O}\left(\frac{e^{-z}}{z^{N+2}}\right) \end{align}

Which concludes the asymptotic expansion.