Limit with Integration - 2

Let $y=(x^{n}+(1-x)^{n})^{\frac{1}{n}}=(1-x)\left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}}$ Now, in $0\leq x \leq \frac{1}{2}$ $0\leq \frac{x}{1-x} \leq 1$ $0\leq \left(\frac{x}{1-x}\right)^{n} \leq 1$ $1\leq \left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$ $(1-x)\leq y \leq 2^{\frac{1}{n}}(1-x)$ $\int_0^{\frac{1}{2}}(1-x)dx \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}}\int_0^{\frac{1}{2}}(1-x)dx$ $\frac{3}{8} \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}} \frac{3}{8}$

for $x$ tending to infinity $2^{\frac{1}{n}}$ tends to $1$ . So, the limit is : $\int_0^{\frac{1}{2}} y\:dx =\frac{3}{8}$

$\lim_{n \to \infty} \int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=2\lim_{n \to \infty} \int^{\frac{1}{2}}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=2\cdot \color{#D61F06}{\frac38}=\boxed{\color{#3D99F6}{\frac{3}{4}}}$

Another possible solution uses Lebesgue's Dominated Convergence Theorem . We define a sequence of functions $f_n(x)$ and rewrite it to: $\begin{aligned} f_n:&& D&:=[0;\:1]\rightarrow\mathbb{R}, &&&&& f_n(x)&:=\left( x^n + (1-x)^n \right)^{\frac{1}{n}}=\max\{x,\:1-x\}\left( 1 + \frac{ \min\{x,\:1-x\}^n }{ \max\{x,\:1-x\}^n } \right)^{\frac{1}{n}} \end{aligned}$ We use the rewritten form of $f_n(x)$ to find an upper and a lower estimate. We obtain its point-wise limit for $n\rightarrow\infty$ via "Squeeze Theorem": $\begin{aligned} x&\in D:&&& 0\leq \max\{x,\:1-x\} &\leq f_n(x)\leq \max\{x,\:1-x\}\cdot 2^{\frac{1}{n}} &&&\Rightarrow &&&&\lim_{n\rightarrow\infty} f_n(x)&=\max\{x,\:1-x\}=:f(x) \end{aligned}$ A slightly sloppier estimation yields a dominating function $g(x)$ of the sequence $f_n(x)\geq 0$ : $\begin{aligned} n&\geq 1: &&& |f_n(x)|=f_n(x)\leq f(x)\cdot 2^{\frac{1}{n}}\leq 2f(x) =:g(x)\text{ integrable over }D \end{aligned}$ The sequence $f_n(x)$ satisfies all prerequisites of the Dominated Convergence Theorem , so we may interchange limit and integration: $\lim_{n\rightarrow\infty}\int_Df_n(x)\:dx=\int_Df(x)\:dx\underset{\text{symmetry}}{=}2\int_0^{\frac{1}{2}} f(x)\:dx=2\int_0^{\frac{1}{2}} 1-x\:dx=2\left[ x - \frac{x^2}{2} \right]_0^{\frac{1}{2}}=2\cdot\frac{3}{8}=\boxed{\frac{3}{4}}$

$I = \int_{0}^{1} \left( x^n + (1-x)^n \right) ^{1/n} \ dx$

Consider the expression as $x$ varies from $0$ to $1$ .

$p(x)=\frac{x}{1-x}$

As $x$ varies from $0$ to $0.5$ , $p \le 1$ and beyond $x=0.5$ , $p > 1$ . Keeping this in mind, the integral can be split as follows:

$I = \int_{0}^{0.5} \left( x^n + (1-x)^n \right) ^{1/n} \ dx + \int_{0.5}^{1} \left( x^n + (1-x)^n \right) ^{1/n} \ dx$ $I = \int_{0}^{0.5} (1-x) \left( 1 + \left(\frac{x}{1-x}\right)^n \right) ^{1/n} \ dx + \int_{0.5}^{1} x \left( 1 + \left(\frac{1-x}{x}\right)^n \right) ^{1/n} \ dx$

Now, since the range of $p(x)$ is established the two integrands can be approximated using a binomial expansion as such:

$I = \int_{0}^{0.5} (1-x) \left( 1 + \frac{1}{n}\left(\frac{x}{1-x}\right)^n \right) \ dx + \int_{0.5}^{1} x \left( 1 + \frac{1}{n}\left(\frac{1-x}{x}\right)^n \right) \ dx$ $I = \int_{0}^{0.5} (1-x) \left( 1 + \frac{\left(p(x)\right)^n}{n} \right) \ dx + \int_{0.5}^{1} x \left( 1 + \frac{1}{n} \left(\frac{1}{p(x)}\right)^n \right) \ dx$

Now, when $0 \le x < 0.5$ then:

$p=\frac{x}{1-x} <1$ $\implies \lim_{n \to \infty} \frac{(p(x))^n}{n} = 0 \ \forall \ 0 \le x < 0.5$

And when $0.5 < x \le 1$ , then:

$\frac{1}{p}=\frac{1-x}{x} <1$ $\implies \lim_{n \to \infty} \frac{(1/p(x))^n}{n} = 0 \ \forall \ 0.5 < x \le 1$

Applying these results to the integrands leads to (for $n \to \infty$ ):

$I = \int_{0}^{0.5} (1-x) \ dx + \int_{0.5}^{1} x \ dx$ $I = \frac{3}{8} + \frac{3}{8} = \boxed{0.75}$