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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?

2 Answers 2

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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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    Can you give a hint without Taylor expansion? I have no idea how to use them... – 2017-02-27
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    @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? – 2017-02-27
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    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. – 2017-02-27
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    @Liviu Are you familiar with the Mean Value Theorem? – 2017-02-27
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    @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. – 2017-02-27
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    @Dr.MV Yes, I am. – 2017-02-27
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    Jack, in the expansion you wrote, $o(x-1)^2$ should be $o(x-1)$. That's not a big deal, but there might be something insidious here. The term in brackets $\left[\cdot\right]$ must be positive, but how does one know that it doesn't approach zero as $n\to \infty$ in such a way such that $\left(\left[\cdot\right]\right)^{1/n}$ does not approach $1$? – 2017-02-27
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    @Dr.MV: typo fixed, thanks. Of course, one has to achieve a sort-of uniform control over the neglected terms, but that is simple through convexity or similar means. – 2017-02-27
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    Jack, you're welcome. And you might consider mentioning something about how we can guarantee that the bracketed term doesn't run away to $0$ too quickly or at all. That is, if we ignored the higher order terms, then $1-\alpha n^2\to -\infty$. So, those "neglected terms" are critically important to the limit. – 2017-02-27