3

Consider the integral $$\int_1^\infty\frac{\exp(-nx)}{x}dx$$

We get:$$\int_1^\infty\frac{\exp(-nx)}{x}dx=n\int_ n^\infty\frac{\exp(-x)}{x}=nE_1(n)$$

My question is, can we approximate this integral with elementary functions for large n?

I've found out that $$\frac {e^{-n}} n> \int_1^\infty\frac{\exp(-nx)}{x}dx$$ As,$$\int_1^\infty\frac{\exp(-nx)}{x}dx<\int_1^\infty{\exp(-nx)}dx$$ Also,$$\int_1^\infty\frac{\exp(-nx)}{x}dx>\sum_{k=1}^{\infty} \dfrac{\exp(-nk)(1-\exp(-n))}{n(k+1)}=\frac{\exp(-n)}{2n}+\mathcal{O}(1)\exp(-2n)$$

Ishan Banerjee
  • 6,459

1 Answers1

3

Integration by parts does the trick here:

$$\begin{align}n \int_n^{\infty} dx \frac{e^{-x}}{x} &=e^{-n} - n \int_n^{\infty} dx \frac{e^{-x}}{x^2}\\&= e^{-n} -\frac{e^{-n}}{n} + 2 n \int_n^{\infty} dx \frac{e^{-x}}{x^3} \end{align}$$

and so on. The full asymptotic expansion is

$$n \int_n^{\infty} dx \frac{e^{-x}}{x} \sim e^{-n} \sum_{k=0}^{\infty} \frac{(-1)^k \, k!}{n^k} \quad (n \to \infty)$$

Note that the sum is a divergent series for a fixed, finite $n$, which means practically, you evaluate the sum for a given $n$ such that the last term used is a minimum in magnitude.

Ron Gordon
  • 138,521
  • @Ishan, you can check when the terms in the series begin to increase in magnitude by solving $k!/n^k = (k+1)!/n^{k+1}$, giving $k = n-1$. So a good approximation is $$n \int_n^{\infty} dx \frac{e^{-x}}{x} \approx e^{-n} \sum_{k=0}^{n-1} \frac{(-1)^k , k!}{n^k}.$$ – Antonio Vargas Jun 14 '13 at 18:41