Csdrhrt

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
$$log x = lim_{h to 0} frac{x^h - 1}h quad text{, or equivalently} quadlog x = lim_{n to infty} n(x^{frac1n} - 1)$$

That is, my solution is a process that involves terms like $~e^{blog b},~$ 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
$$ M_p(a_1,ldots,a_m) = left(frac{1}{m}sum_{k=1}^{m}a_k^pright)^{frac{1}{p}},$$
then $M_p$ is an increasing function of $p$ and
$$ lim_{pto 0^+} M_p(a_1,ldots,a_m)= GM(a_1,ldots,a_m) = left(prod_{k=1}^{m}a_kright)^{frac{1}{m}}$$
by the continuity and concavity of the logarithm function.

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

$$ lim_{pto 0^+}left(int_{0}^{1}f(x)^p,dxright)^{frac{1}{p}}=expint_{0}^{1}log f(x),dx.$$

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

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

$newcommand{bbx}[1]{,bbox[8px,border:1px groove navy]{{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
&lim_{n to infty}braces{int_{0}^{1}bracks{bx + apars{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^{+}}exppars{-lnpars{b - a} + lnpars{int_{a}^{b}x^{epsilon},dd x} over epsilon}
\[5mm] & =
lim_{epsilon to 0^{+}}exppars{-lnpars{b - a} +
lnpars{b^{epsilon + 1} - a^{epsilon + 1}} - lnpars{epsilon + 1} over epsilon}
\[5mm] & =
lim_{epsilon to 0^{+}}exppars{
{b^{epsilon + 1}lnpars{b} - a^{epsilon + 1}lnpars{a} over b^{epsilon + 1} - a^{epsilon + 1}} - {1 over epsilon + 1}}qquadqquad
pars{~L'Hhat{o}pital Rule~}
\[5mm] & =
exppars{{blnpars{b} - alnpars{a} over b - a} - 1} =
bbx{ds{{1 over expo{}},pars{b^{b} over a^{a}}^{1/pars{b - a}}}}
end{align}