TITLE: Limit with integral and power
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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 $C\cdot\delta\left(t-\frac{1}{e}\right)$.