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Find the limit,

$$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$

My try:

$$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$

Now taking the limit I get that,

$$\frac{1}{4}

But, how can I get the exact answer!!

This is Problem 11941 from the American Mathematical Monthly.

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    this was discussed on this site not so long ago... – 2017-01-06
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    @tired can you show me the question id? – 2017-01-06
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    i can't find it at the moment---:(. Nevertheless the limit should be $3/4$ which is suggested by some simple plots – 2017-01-06
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    This is twice $$\int_{1/2}^1x\cdot\sqrt[n]{1+(x^{-1}-1)^n}\cdot dx$$ What can you say about the behaviour of $$(x^{-1}-1)^n$$ and about the behaviour of $$\sqrt[n]{1+(x^{-1}-1)^n}$$ when $n\to\frac12$, for $x$ in $(\frac12,1)$? – 2017-01-06
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    @Did the function is bounded by 1, isnt it? – 2017-01-06

5 Answers 5

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Substitute 1/2+y for x, y going from -1/2 to 1/2. This a symmetric integral so the positve and negative intervals contribute equally and it suffices to consider only the positve interval. As n goes to infinity (1/2-y)^n is neglible compared to (1/2+y)^n. Hence the root approaches 1/2+ y and the integral approaches 2*Int((1/2+y)*dy) = 1/2+1/4 = 3/4

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    The $n^{th}$ root in your integral converges to $1$ for $1\le y\le2$, and $y-1$ for $y>2$, so this isn't quite right. – 2017-01-01
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    But the answer is 3/4 – 2017-01-01
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Hint/Intuition:

For $x,y\ge0, \lim_{n\to\infty} \left(x^n+y^n\right)^{1/n}=\max\{x,y\}$.

Alternative Approach:

$$I=\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx=2\int_{0}^{1/2}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$

$$=2\int_0^{1/2}(1-x)\left(1+\left(\frac{x}{1-x}\right)^n\right)^{1/n}$$

Let $u=\frac{x}{1-x}\implies dx=\frac{du}{(1+u)^2}$

$$I=2\int_0^1\frac{1}{1+u}(1+u^n)^{1/n}\frac{du}{(1+u)^2}$$

Note that for $y>0, 1<(1+y)^{1/n}<1+\frac{y}{n}$, so:

$$2\int_0^1\frac{du}{(1+u)^3}\le I \le 2\int_0^1\left(1+\frac{u^n}{n}\right)\frac{du}{(1+u)^3}$$

Now:

$$0\le I - 2\int_0^1\frac{du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{u^n\,du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4n}$$

So, the integral converges to $2\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4}$

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    How is it straight forward. There is an integration before that – 2017-01-01
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    Ah, that's sort of true. Well, the idea is that the interior function is going to look more and more like $\max\{x,1-x\}$ as $n$ increases, and by being a bit more precise with this approximation, you get that the integral converges as we'd like it to. – 2017-01-01
4

Use the better estimate $$\int_{0}^{\frac{1}{2}}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}x \,dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}\,dx< \int_{0}^{\frac{1}{2}}2^{1/n}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}2^{1/n}x\,dx$$

Because $2^{1/n} \to 1,$ the squeeze theorem tells us the limit is $\int_{0}^{\frac{1}{2}}(1-x)\,dx+ \int_{\frac{1}{2}}^{1}x \,dx = 3/4.$

4

From norm equivalences one remembers $$ \max(a,b)\le\sqrt[n]{a^n+b^n}\le\max(a,b)\sqrt[n]2 $$ which can be used to get upper and lower bounds for the integral, $$ \frac34\le\int_0^1\sqrt[n]{x^n+(1-x)^n}\,dx\le \frac34\;\sqrt[n]2 $$ which nicely gives the limit as $\frac34$.

3

Note that $$\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx\leq\int_{0}^{1}1dx=1 $$ so from the Dominated Convergence Theorem we get $$\lim_{n\rightarrow\infty}\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx=\int_{0}^{1}\lim_{n\rightarrow\infty}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx $$ and $$\lim_{n\rightarrow\infty}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}=\exp\left(\lim_{n\rightarrow\infty}\frac{\log\left(x^{n}+\left(1-x\right)^{n}\right)}{n}\right) $$ $$=\exp\left(\lim_{n\rightarrow\infty}\frac{x^{n}\log\left(x\right)+\left(1-x\right)^{n}\log\left(1-x\right)}{x^{n}+\left(1-x\right)^{n}}\right)=\max\left\{ x,1-x\right\} $$ hence $$\lim_{n\rightarrow\infty}\int_{0}^{1}\left(x^{n}+\left(1-x\right)^{n}\right)^{1/n}dx=\int_{0}^{1}\max\left\{ x,1-x\right\} dx $$ $$=\int_{0}^{1/2}\left(1-x\right)dx+\int_{1/2}^{1}xdx=\color{red}{\frac{3}{4}}.$$