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$$L = \displaystyle \lim_{{n\to\infty}}\sqrt[n]{\int_0^1{(\arctan(1 + x^n))^n dx}}$$

I started with $0 \le x \le 1$ and I got to $\frac{\pi}{4} \le L \le \arctan 2$ I don't really know if that helps ... The answer is $\arctan 2$.

Also, $$\displaystyle \lim_{{n\to\infty}}\sqrt[n]{\int_0^1{(1+x^n)^n dx}}$$

I saw this as

$$e^{\lim _{n\to \infty }\frac{1}{n}\ln \left(\int _0^1\left(1+x^n\right)^n\:dx\right)}$$

Can I apply L'hospital somehow here?

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Here's a generalization that is no harder than the specific problem.

Thm: Suppose $f$ is continuous, increasing, and nonnegative on $[0,1].$ Define $I_n = \int_0^1 f(x^n)^n\,dx.$ Then

$$\lim_{n\to \infty} I_n^{1/n} = f(1).$$

With $f(x) = \arctan (1+x),$ the theorem shows the limit in the given problem is $f(1)=\arctan 2.$

Proof of the theorem: Since $I_n \le f(1)^n,$ we have $I_n^{1/n} \le f(1)$ for all $n.$ To get a lower bound, make the change of variables $x=y^{1/n}$ to see

$$I_n =\int_0^1 f(y)^n \, (1/n)y^{1/n-1}\,dy.$$

Let $a\in (0,1).$ Then

$$\frac{f(a)^n}{n}\int_a^1 y^{1/n-1}\,dy < I_n.$$

Because $1\le y^{1/n-1}$ on $(0,1]$ for all $n,$ we then have

$$\frac{f(a)^n}{n}(1-a) < I_n.$$

Thus

$$f(a)\left (\frac{1-a}{n}\right )^{1/n} < I_n^{1/n}.$$

Now as $n\to \infty,$ both $(1-a)^{1/n}, n^{1/n} \to 1.$ Therefore

$$f(a) \le \liminf_{n\to \infty} I_n^{1/n}.$$

This is true for each $a\in (0,1);$ therefore it holds in the limit as $a\to 1^-.$ By the continuity of $f,$ we get

$$ f(1) \le \liminf_{n\to \infty} I_n^{1/n}.$$

Recalling our easy upper bound, we thus have

$$\limsup_{n\to\infty} I_n^{1/n} \le f(1) \le \liminf_{n\to \infty} I_n^{1/n} \le \limsup_{n\to\infty} I_n^{1/n}.$$

All expressions here are thus equal, proving the theorem.

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Since $\arctan(2)>1$, most of the mass of the integral $\int_{0}^{1}\arctan(1+x^n)^n\,dx $ is concentrated in a left neighbourhood of $x=1$. By considering the Taylor series of $\arctan(1+x^n)^n$ around $x=1$ we get $$ \arctan(1+x^n)^n = \arctan(2)^n\left[1+\frac{n^2}{5\arctan 2}(x-1)+o(x-1)\right] $$ hence $\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(1+x^n)^n\,dx }=\arctan(2)$, since we are dealing with convex functions, ensuring that

$$\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(1+x^n)^n\,dx }=\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(2)^n \exp\left(\frac{n^2}{5\arctan 2}(x-1)\right)\,dx }.$$

By a similar argument (a simplified version of Laplace's method) $$ \lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}(1+x^n)^n\,dx}=2.$$ Here $(1+x^n)^n$ is even more than convex, it is log-convex.

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  • $\begingroup$ Can you give a hint without Taylor expansion? I have no idea how to use them... $\endgroup$ – Liviu Feb 27 '17 at 16:31
  • $\begingroup$ @Liviu: so you are approaching parametric integrals and the dominated convergence theorem without having any clue about Taylor expansions? What kind of Calculus course are you following? $\endgroup$ – Jack D'Aurizio Feb 27 '17 at 16:53
  • $\begingroup$ I don't even know about dominating convergence, even though I've heard about it. I'm from Romania, in 12th grade, and this should be some problems similar to the ones that can be in the test for University (computer science). So I have only studied some properties of the integrals (about monotony, sign, periodicity, Riemann sums, etc), but not what you have exposed in the comment or answer. $\endgroup$ – Liviu Feb 27 '17 at 17:04
  • $\begingroup$ @Liviu Are you familiar with the Mean Value Theorem? $\endgroup$ – Mark Viola Feb 27 '17 at 17:08
  • $\begingroup$ @Liviu: all right, in such a case try to figure out the shape of the graph of $(1+x^n)^n$ over $(0,1)$ as $n\to +\infty$. In a right neighbourhood of $x=0$ such function is very close to $1$, while in a right neighbourhood of $x=1$ it has a large value ($\approx 2^n$) and a positive derivative. Try to approximate such function trough simpler functions and derive the result by squeezing. $\endgroup$ – Jack D'Aurizio Feb 27 '17 at 17:08

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