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Let $x$ be a positive real number and $f(x):=\lim_{\epsilon\to0}\int_\epsilon^{\infty} \dfrac{e^{xt}}{t^t} \, dt $. How fast does this function grow ? In other words can we find a good asymptote for $f(x)$ as $x$ goes to $+\infty$ ?

Can we show one of these two limits converges to a constant :

A) $\lim_{x\to+\infty} \dfrac{\ln(f(x))}{P(x)} $

B) $\lim_{x\to+\infty} \dfrac{f(x)}{P(x)} $

For some polynomial $P(x)$ ?

I know $f(z)$ is an entire function , so I tried using Taylor series.

However the derivatives of $f$ are similar looking and Hence I do not know their growth rate either !?

$$\frac{d f(x)}{d x^k} = \lim_{\epsilon\to0}\int_\epsilon^\infty \frac{e^{xt}}{t^{t-k}}\,dt.$$

Since by Taylor's theorem I need the derivatives of $f(x)$, so I am stuck on how to prove any growth rate or limit.

I considered replacing the integral with an infinite sum but that did not work for me.

I assume one way is to use contour integrals but I'm not sure how that would work.

Martin Sleziak
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mick
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  • Why not just write the lower bound for that integral as zero, instead of using that limit in front? – Ethan Splaver Dec 27 '13 at 22:05
  • @Ethan because $t^t$ is not well defined for $t=0$. – mick Dec 27 '13 at 22:09
  • Yes but neither is it for your upper bound at infinity, why write half of it as an improper integral and then use a limit for the other half? – Ethan Splaver Dec 27 '13 at 22:10
  • It is well defined for t=+oo. $\frac{1}{t^t}=0$ for t= + oo. @Ethan – mick Dec 27 '13 at 22:15
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    Mick, $\lim_{t\to 0} \frac{1}{t^t}=1$ in addition to $\lim_{t\to \infty}\frac{1}{t^t}=0$ – Ethan Splaver Dec 27 '13 at 22:16
  • It is, if $t>0$, as is clearly the case here. – Lucian Dec 27 '13 at 22:18
  • @Ethan I know. I know. – mick Dec 27 '13 at 22:18
  • The Laplace method yields $$f(x) \sim \sqrt{2\pi} \exp\left(e^{x-1}+(x-1)/2\right),$$ though I haven't verified the details. I'll try to write up an answer soon but I'm currently travelling so I don't know when that will be. – Antonio Vargas Dec 28 '13 at 10:27
  • It is easy to show that this function must grow faster than the gamma function, so the limit B) does not exist. If Antonio is correct then the limit A) also does not exist. – mick Jan 03 '14 at 20:19

1 Answers1

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First we can rewrite the integral as

$$ f(x) = \int_0^\infty \exp\left\{x t - t\log t\right\}\,dt. $$

The function $g(t) = xt - t\log t$ has a maximum when $t = e^{x-1}$. This suggests the change of variables

$$ t = e^{x-1}(1+s), $$

which transforms our integral into

$$ f(x) = \exp\left\{e^{x-1} + x - 1\right\} \int_{-1}^{\infty} \exp\left\{e^{x-1} \Bigl[s - s\log(1+s) - \log(1+s)\Bigr]\right\}\,ds. $$

(I've found this trick to be pretty useful. I learned it while reading about the asymptotics for the Gamma function and I also used it in this answer.) Near $s=0$ we have

$$ s - s\log(1+s) - \log(1+s) \sim -\frac{1}{2}s^2, $$

so the Laplace method yields immediately

$$ \begin{align} &\int_{-1}^{\infty} \exp\left\{e^{x-1} \Bigl[s - s\log(1+s) - \log(1+s)\Bigr]\right\}\,ds \\ &\qquad \sim \int_{-\infty}^{\infty} \exp\left\{-\frac{1}{2}e^{x-1}s^2\right\}\,ds \\ &\qquad = \sqrt{2\pi} e^{-(x-1)/2} \end{align} $$

as $x \to \infty$. Thus

$$ f(x) \sim \sqrt{2\pi} \exp\left\{e^{x-1} + \frac{x-1}{2}\right\} $$

as $x \to \infty$. If one desired they could obtain more terms of the asymptotic expansion using the method in this answer. For example, by including the next term we get

$$ f(x) = \sqrt{2\pi} \exp\left\{e^{x-1} + \frac{x-1}{2}\right\} \left[1 + \frac{5}{24}e^{1-x} + O\left(e^{-2x}\right)\right] $$

as $x \to \infty$.

Community
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Antonio Vargas
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  • Im sorry I missed this answer. Thank you. Although I did not yet check the correctness of it completely it seems valid. Perhaps a bit " fast " explained for beginners. Those links are intresting and helpfull though. I need to practice on these " techniques ". Its new to me. – mick Jan 28 '14 at 20:45
  • You're welcome! The basic idea is that integrals of the form $\int_{-a}^{b} e^{xf(s)},ds$, where $f(s)$ has a maximum at $t=0$, can be approximated by replacing $f(s)$ with the first two relevant terms of its Maclaurin series and replacing the integral $\int_{-a}^{b}$ with $\int_{-\infty}^{\infty}$. In the proof of the Laplace method it is shown that doing so will only introduce errors which are exponentially smaller than the original integral as $x \to \infty$. In this case we had $f(s) = s-s\log(1+s)-\log(1+s) = 0-s^2/2 + O(s^3)$, so we replaced $f(s)$ with $-s^2/2$. – Antonio Vargas Jan 28 '14 at 21:36
  • A good reference for the Laplace method is de Bruijn's Asymptotic Methods in Analysis. – Antonio Vargas Jan 28 '14 at 21:37