Limit of $\left\{ \int_{0}^{1} [bx + a(1 - x) ]^{\frac1n} dx \right\}^n$ as $n \to \infty$

This is just the continuous analogue of a well-known fact:

If $a_1,a_2,\ldots,a_m$ are non-negative numbers and the mean of order $p>0$ is defined as $$ M_p(a_1,\ldots,a_m) = \left(\frac{1}{m}\sum_{k=1}^{m}a_k^p\right)^{\frac{1}{p}},$$ then $M_p$ is an increasing function of $p$ and $$ \lim_{p\to 0^+} M_p(a_1,\ldots,a_m)= GM(a_1,\ldots,a_m) = \left(\prod_{k=1}^{m}a_k\right)^{\frac{1}{m}}$$ by the continuity and concavity of the logarithm function.

In particular, if $f(x)$ is a non-negative and continuous function over the interval $(0,1)$,

$$ \lim_{p\to 0^+}\left(\int_{0}^{1}f(x)^p\,dx\right)^{\frac{1}{p}}=\exp\int_{0}^{1}\log f(x)\,dx.$$

The last identity gives that for any $a,b>0$,

$$ \begin{eqnarray*}\lim_{n\to +\infty}\left(\int_{0}^{1}(bx+a(1-x))^{\frac{1}{n}}\,dx\right)^n &=& \exp\int_{0}^{1}\log(bx+a(1-x))\,dx\\&=&\exp\left[\frac{1}{b-a}\int_{a}^{b}\log(x)\,dx\right]\\&=&\exp\left[-1+\frac{b\log b-a\log a}{b-a}\right]\\&=&\color{red}{\frac{1}{e}\left(\frac{b^b}{a^a}\right)^{\frac{1}{b-a}}} \end{eqnarray*}$$ as claimed.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\braces{\int_{0}^{1}\bracks{bx + a\pars{1 - x}}^{1/n} \,\dd x}^{n} = \lim_{n \to \infty}\bracks{{1 \over b - a}\int_{a}^{b}x^{1/n} \,\dd x}^{n} \\[5mm] &= \lim_{\epsilon \to 0^{+}}\exp\pars{-\ln\pars{b - a} + \ln\pars{\int_{a}^{b}x^{\epsilon}\,\dd x} \over \epsilon} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\exp\pars{-\ln\pars{b - a} + \ln\pars{b^{\epsilon + 1} - a^{\epsilon + 1}} - \ln\pars{\epsilon + 1} \over \epsilon} \\[5mm] & = \lim_{\epsilon \to 0^{+}}\exp\pars{ {b^{\epsilon + 1}\ln\pars{b} - a^{\epsilon + 1}\ln\pars{a} \over b^{\epsilon + 1} - a^{\epsilon + 1}} - {1 \over \epsilon + 1}}\qquad\qquad \pars{~L'H\hat{o}pital\ Rule~} \\[5mm] & = \exp\pars{{b\ln\pars{b} - a\ln\pars{a} \over b - a} - 1} = \bbx{\ds{{1 \over \expo{}}\,\pars{b^{b} \over a^{a}}^{1/\pars{b - a}}}} \end{align}