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

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

Just to be clear, 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.

I actually have been able to solve it as I happen to know a particular way of viewing the logarithm function as a limit $$\log x = \lim_{h \to 0} \frac{x^h - 1}h \quad \text{, or equivalently} \quad\log x = \lim_{n \to \infty} n(x^{\frac1n} - 1)$$

That is, my solution is a process that involves terms like $~e^{b\log b},~$ where I have to obtain $~\log b~$ first as a limit, and I can say that it is $~b^b~$ at the very last step only after another limit.

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

In short, I have a solution which I don't like, and I believe there are better ones.

Any ideas? Thank you.

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

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}

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.
