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I am trying to show that

$$\lim {\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}=\frac 1 2 $$

Now, this can be done as follows. Using $x\mapsto x^{-1}$ we get that

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} =\int\limits_1^\infty {{{nx^{n-1}} \over {1 + {x^n}}}} \frac{dx}{nx}$$

Upon integrating by parts we get

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} $$

or $$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} $$ since $\int_1^\infty x^{-2}dx=1$.

Using $x\mapsto x^{-1}$ once again on the RHS, we get

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_0^1 {{{\log \left( {1 + {x^n}} \right) - n\log x - n} \over n}dx} $$

But since $\int_0^1 \log x=-1=\int_0^1 dx$

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)dx} $$

Now, since for $x\geq 0$, $\log(1+x)\leq x$, we get $$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} \le 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {{x^n}dx} = 1 - {{\log 2} \over n} + {1 \over {n\left( {n + 1} \right)}}$$ which means

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + O\left( {{1 \over {{n^2}}}} \right)$$

Is this enough to conclude that

$${\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n} = {e^{ - \log 2}} = {1 \over 2}\text{ ? }$$

If so, how?

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  • $\begingroup$ I am all in favor of using \limits in titles, but this one was too large! :-) $\endgroup$ – Asaf Karagila Feb 20 '13 at 22:00
  • $\begingroup$ @AsafKaragila Hehe, no problem. $\endgroup$ – Pedro Tamaroff Feb 20 '13 at 22:01
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Your question can be recast as: if $$ a_n=1+\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr), $$ is it true that $$ \lim_{n\to\infty}a_n^n=e^a? $$ The answer is yes. Since $\log(1+x)=x+O(x^2)$ as $x\to0$, we have $$ \log a_n=\frac{a}{n}+O\Bigl(\frac{1}{n^2}\Bigr). $$ Then $$ \lim_{n\to\infty}a_n^n=\lim_{n\to\infty}e^{n\log a_n}=e^a. $$

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    $\begingroup$ No, it will not work with $O(1/n)$, since you need the limit of $n\log a_n$. It will work however with $o(1/n)$. $\endgroup$ – Julián Aguirre Feb 20 '13 at 22:22
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    $\begingroup$ To restate Julián's result: if $c_n\to c$, then $\left(1+{c_n\over n}\right)^n\to e^c$. This even works for complex numbers $c_n$. $\endgroup$ – user940 Feb 20 '13 at 23:39
  • $\begingroup$ Oh, right, the $n$ would make things go awry. Maybe I failed to understand how you work when getting $O$ inside $O$s? $\endgroup$ – Pedro Tamaroff Feb 20 '13 at 23:39
  • $\begingroup$ @ByronSchmuland I don't think that is what Julián states. (?) $\endgroup$ – Pedro Tamaroff Feb 20 '13 at 23:40
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    $\begingroup$ @ByronSchmuland: your restatement is essentially my comment about equicontinuous functions in my answer since $\left(1+\frac xn\right)^n$ is an equicontinuous sequence on compact sets. $\endgroup$ – robjohn Feb 21 '13 at 1:07
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$$ \begin{align} \hspace{-2cm}\lim_{n\to\infty}\left(\int_0^1\frac{\mathrm{d}x}{1+x^n}\right)^n &=\lim_{n\to\infty}\left(1-\int_0^1\frac{x^n\mathrm{d}x}{1+x^n}\right)^n\\ &=\lim_{n\to\infty}\left(1-\frac1n\int_0^1\frac{x\mathrm{d}x^n}{1+x^n}\right)^n\\ &=\lim_{n\to\infty}\left(1-\left[\frac1nx\log(1+x^n)\right]_0^1+\frac1n\int_0^1\log(1+x^n)\mathrm{d}x\right)^n\\ &=\lim_{n\to\infty}\left(1-\frac1n\log(2)+\frac1n\int_0^1\log(1+x^n)\mathrm{d}x\right)^n\\ &=\lim_{n\to\infty}\left(1-\frac1n\log(2)+\frac1n\int_0^1O(x^n)\,\mathrm{d}x\right)^n\tag{$\log(1+x^n)\le x^n$}\\ &=\lim_{n\to\infty}\left(1-\frac1n\log(2)+O\left(\frac1{n^2}\right)\right)^n\\ &=\lim_{n\to\infty}\left(1-\frac1n\left(\log(2)+O\left(\frac1n\right)\right)\right)^n\\ &=e^{-\log(2)}\\ \end{align} $$ Note that $\lim\limits_{n\to\infty}f_n(x_n)=\lim\limits_{n\to\infty}f_n\left(\lim\limits_{n\to\infty}x_n\right)$ when $\{f_n\}$ are equicontinuous at $\lim\limits_{n\to\infty}x_n$.

In this case $$ f_n(x)=\left(1-\frac xn\right)^n $$ and $$ x_n=\log(2)+O\left(\frac1n\right) $$

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  • $\begingroup$ The statement on equicontinuity is worth reading about. $\endgroup$ – Pedro Tamaroff Feb 21 '13 at 1:10
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$$\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}\right)^{n} = e^{\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}-1\right)n}$$

But $$\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0} \frac{dx}{1+x^{n}}-1\right)n = -\lim\limits_{n\rightarrow\infty}\left(\int^{1}_{0}x[\ln(1+x^{n})]'dx\right)=$$

$$ -\lim\limits_{n\rightarrow\infty}x\ln(1+x^{n})|^{1}_{0} + \lim\limits_{n\rightarrow\infty}\int^{1}_{0}\ln(1+x^{n})dx=-\ln 2$$

because $\ln(1+x^{n}) \leq x^{n}$ it involves $$\lim\limits_{n\rightarrow\infty}\int^{1}_{0}\ln(1+x^{n})dx = 0$$

The conclusion is immediate.

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  • $\begingroup$ Did you read the other work, both by myself and the answerers? I'm not saying your answer is not helpful, but it really doesn't add anything new of what has already been said. $\endgroup$ – Pedro Tamaroff May 4 '13 at 22:42
  • $\begingroup$ Also, how do you justify the first step equation? $\endgroup$ – Pedro Tamaroff May 4 '13 at 22:46

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