6

Evaluate the limit for $0 < a < b$, $$\lim_{n \to \infty} \left\{ \int_{0}^{1} [bx + a(1 - x) ]^{\frac1n} dx \right\}^n$$

Note that the exponent $~1/n~$ is inside the integration on the entire integrand $~a + (b-a)x,~$ and the exponent $n$ is outside on the definite integral.

One solution is available given this particular definition of the logarithm function as a limit: $$\log x = \lim_{h \to 0} \frac{x^h - 1}h \quad \text{, or equivalently} \quad\log x = \lim_{n \to \infty} n(x^{\frac1n} - 1)$$

This solution is an algebraic maneuver that involves terms like $~e^{b\log b},~$ where one first obtains $~\log b~$ as a limit, and then only after another limit can one arrive at $~b^b~$ eventually.

My question is this: how does one evaluate the limit more directly without this seemingly redundant path of log on the exponent?

In short, there's an approach at hand that is unsatisfactory, and I believe there are better ones.

As a reference, below is the detailed steps of the circuitous solution outlined above:

The definite integral evaluates to $$\frac1{b - a} \frac1{ 1 + \frac1n } \left[ a + (b-a)x \right]^{ 1 + \frac1n } \Bigg|_{0}^{1} = \frac{ b^{1 + \frac1n} - a^{1 + \frac1n} }{b - a} \frac1{ 1 + \frac1n }$$ Thus the whole expression becomes $$ \begin{align} &\lim_{n \to \infty} \left\{ \frac{ b^{1 + \frac1n} - a^{1 + \frac1n} }{b - a} \frac1{ 1 + \frac1n } \right\}^{n} \\ &= \frac1e \lim_{n \to \infty} \left\{ 1 + \frac{ b^{1 + \frac1n} - a^{1 + \frac1n} - (b - a) }{b - a} \right\}^{n} \\ &= \frac1e \lim_{n \to \infty} \left\{ 1 + \frac{ n \left( b^{1 + \frac1n} - b \right) - n \left( a^{1 + \frac1n} - a \right) }{ n (b - a) } \right\}^{n} \end{align} $$ all the limits exist so I'm just gonna keep writing in this non-rigorous way $$ \begin{align} &= \frac1e \lim_{n \to \infty} \left\{ 1 + \frac1n \frac{ b \log b - a \log a }{ b - a } \right\}^{n} \\ &= \frac1e \cdot e^{ \frac1{b-a} \left( b \log b - a \log a \right)} \\ &= \frac1e \left( \frac{ b^b }{ a^a } \right)^{ \frac1{b-a}} \end{align}$$

Lee David Chung Lin
  • 7,043

2 Answers2

7

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.

Jack D'Aurizio
  • 353,855
7

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
  • 89,464