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I'm trying to calculate this limit:

$$\lim_{x\to\infty} \left(\int_0^1 t^{-tx} dt\right)^{\frac1x}$$

I tried the squeezing idea without success.

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    If $f$ is continuous or just integrable , $\lim_{n \to \infty} \left(\int_0^1 |f(x)|^n\, dx\right)^{1/n} = \sup_{x \in [0,1]} f(x)$ is a well known result but you can arrive at the conclusion with a squeezing argument as hinted. – 2017-02-23

2 Answers 2

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A maximum for $t^{-t}$ on $[0,1]$ is attained at $t = e^{-1}.$

With $0 < t < 1,$ we have

$$t^{-tx} = \exp(-t \ln t)^x \leqslant \exp(e^{-1})^x \\ \implies \left(\int_0^1 t^{-tx} \, dt\right)^{1/x} \leqslant \exp(e^{-1}).$$

Using continuity, for any small $\epsilon$, there is an interval $[e^{-1} - \delta, e^{-1} + \delta] \subset [0,1]$ where $t^{-t} > \exp(e^{-1}) - \epsilon$ and

$$\left(\int_0^1 t^{-tx} \, dt\right)^{1/x} \geqslant \left(\int_{\exp(e^{-1}) - \delta}^{\exp(e^{-1}) + \delta} t^{-tx} \, dt\right)^{1/x} > (\exp(e^{-1})- \epsilon)(2 \delta)^{1/x}. $$

Since $(2\delta)^{1/x} \to 1$ as $x \to \infty$ and $\epsilon$ can be arbitrarily small, the squeeze theorem yields

$$\lim_{x \to \infty}\left(\int_0^1 t^{-tx} \, dt\right)^{1/x} = \exp(e^{-1})$$

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    Perfect use of squeeze +1 – 2017-02-24
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For any large $x$ we have $$ \int_{0}^{1}t^{-tx}\,dt = \int_{0}^{1}\exp\left(-x t\log t\right)\,dt = \sum_{n\geq 0}\frac{x^n}{n!}\int_{0}^{1}t^n(-\log t)^n\,dt =\sum_{n\geq 0}\frac{x^n}{n!(n+1)^{n+1}}$$ If we call this entire function $f(x)$, the wanted limit equals $$\exp\lim_{x\to +\infty}\frac{\log f(x)}{x} \stackrel{dH}{=} \exp\lim_{x\to +\infty}\frac{f'(x)}{f(x)}=\exp\lim_{x\to +\infty}\frac{\int_{0}^{1}(-t\log t)t^{-tx}\,dt}{\int_{0}^{1}t^{-tx}\,dt}$$ that is $\color{red}{e^{1/e}}$ since $t^{-tx}$ converges in distribution to $C\cdot\delta\left(t-\frac{1}{e}\right)$.