Limit with integral and power 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.
 A: 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
$$\lim_{x \to \infty}\left(\int_0^1 t^{-tx} \, dt\right)^{1/x} = \exp(e^{-1})$$
A: 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 $C\cdot\delta\left(t-\frac{1}{e}\right)$.
