5

Evaluate the following limit: \begin{equation} \lim\limits_{t \rightarrow \infty}{(te^t\int\limits_{t}^{\infty}\frac{e^{-s}}{s}ds) := I} \end{equation}

That's the Exponential integral, which is not an elementary function. Considering $ \frac{1}{s} < \frac{1}{t} $ we get $ I \le 1 $, and my intuition says it's actually the answer ($ \lim{I} = 1 $), but I can't prove that $ I \ge 1 $. Am I right?

2 Answers2

7

$$ \begin{align} \lim_{t\to\infty}te^t\int_t^\infty\frac{e^{-s}}{s}\,\mathrm{d}s &=\lim_{t\to\infty}t\int_0^\infty\frac{e^{-s}}{t+s}\,\mathrm{d}s&&s\mapsto t+s\\ &=\lim_{t\to\infty}\int_0^\infty\frac{e^{-s}}{1+s/t}\,\mathrm{d}s\\ &=\int_0^\infty e^{-s}\,\mathrm{d}s&&\begin{array}{}\text{Dominated}\\ \text{Convergence}\end{array}\\[8pt] &=1 \end{align} $$

If an asymptotic expansion is desired, we can use the fact that $\int_0^\infty s^n\,e^{-s}\,\mathrm{d}s=n!$ to get $$ \begin{align} te^t\int_t^\infty\frac{e^{-s}}{s}\,\mathrm{d}s &=\int_0^\infty\frac{e^{-s}}{1+s/t}\,\mathrm{d}s\\ &=\int_0^\infty\left(1-\frac{s}{t}+\frac{s^2}{t^2}-\frac{s^3}{t^3}+\dots\right)\,e^{-s}\,\mathrm{d}s\\ &=0!-\frac{1!}{t}+\frac{2!}{t^2}-\frac{3!}{t^3}+\dots \end{align} $$ As with most asymptotic expansions, this converges nowhere.

robjohn
  • 345,667
3

Integrate by parts:

$$\begin{align}t e^{t} \int_t^{\infty} ds \frac{e^{-s}}{s} &= t e^t \left [ -\frac{e^{-s}}{s}\right]_{t}^{\infty} - t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^2}\\ &= 1-t e^t \left [ -\frac{e^{-s}}{s^2}\right]_{t}^{\infty} + 2 t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^3}\\ &= 1 -\frac{1}{t} + 2 t e^t \left [ -\frac{e^{-s}}{s^3}\right]_{t}^{\infty} - 6 t e^t \int_t^{\infty} ds \frac{e^{-s}}{s^4} \end{align}$$

etc. Note that repeating integration by parts will produce terms in increasing powers of $1/t$. You may then show the limit as being $1$.

robjohn
  • 345,667
Ron Gordon
  • 138,521
  • Looks very good, thanks. But is it formal enough? Although each integral tends to zero, we have an infinite number of them. –  Aug 19 '13 at 03:07
  • @Peregrino: as is, likely not, this is more intuitive. But it certainly can form the basis of something more formal. It should be clear that, while this process is carried out ad infinitum, you may show that each term is less than the previous one as $t \to 0$, so the limit holds. – Ron Gordon Aug 19 '13 at 03:09
  • 1
    I think I can improve your solution, we only need the first line. Considering $1/s \le 1/t$, the last integral in the first line tends to zero, right? –  Aug 19 '13 at 03:12
  • @Peregrino: sure, why not. It's a pretty straightforward integral for asymptotic analysis. – Ron Gordon Aug 19 '13 at 03:17