TITLE: Evaluate $\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$
QUESTION [18 upvotes]: Evaluate
$$\lim_{n \to \infty} \int_{0}^1 [x^n + (1-x)^n ]^{1/n} \ \mathrm{d}x$$

I simplified the limit to
$$\dfrac{1}{2}\lim_{n \to \infty} \int_{0}^{1/2} \left[\left(\frac{1}{2}+x\right)^n + \left(\frac{1}{2}-x\right)^n \right]^{1/n} \ \mathrm{d}x$$
using properties of definite integrals, but I can't solve further.
Any help will be appreciated.
Thanks.

REPLY [2 votes]: It's a little easier to let $x=(1+u)/2$ with $-1\le u\le1$. We have
$$\int_0^1(x^n+(1-x)^n)^{1/n}dx={1\over2}\int_{-1}^1\left(\left(1+u\over2\right)^n+\left(1-u\over2\right)^n \right)^{1/n}du=\int_0^1\left(1+u\over2\right)\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}du$$
where we did two things in the second step: we used the symmetry between $1+u$ and $1-u$ to replace ${1\over2}\int_{-1}^1$ with $\int_0^1$, and we factored out the $\left(1+u\over2\right)^n$.  What you do from here on out depends in part on how finicky you need to be about moving the limit across the integral sign. But since for $0\le u\le1$ we have
$$1\le\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}\le\left(1+\left(1-0\over1+0\right)^n \right)^{1/n}=2^{1/n}$$
we see that
$$\int_0^1\left(1+u\over2\right)du\le\int_0^1\left(1+u\over2\right)\left(1+\left(1-u\over1+u\right)^n \right)^{1/n}du\le2^{1/n}\int_0^1\left(1+u\over2\right)du\to\int_0^1\left(1+u\over2\right)du$$
so the Squeeze Theorem tell us
$$\lim_{n\to\infty}\int_0^1(x^n+(1-x)^n)^{1/n}dx=\int_0^1\left(1+u\over2\right)du={(1+u)^2\over4}\big|_0^1=1-{1\over4}={3\over4}$$